9: Large-Sample Tests of Hypotheses 9.1

a The critical value that separates the rejection and nonrejection regions for a right-tailed test based on a Dstatistic will be a value of D (called D! Ñ such that P(D ž D! Ñ œ ! œ Þ ! " Þ That is, DÞ!" œ # Þ $ $ (see Figure *Þ" ). The null hypothesis H! will be rejected if D ž # Þ $ $ Þ

Fi g u re 9.1

α = .01 z 2 .33

0

Re j ect H 0

b For a two-tailed test with ! œ Þ ! & , the critical value for the rejection region cuts off !Î # œ Þ ! # & in the two tails of the D distributionin Figure *Þ# , so that DÞ!#& œ"Þ*'Þ The null hypothesis H! will be rejected if D ž " Þ * ' or D • • "Þ*' (which you can also write as lDl ž " Þ * ') Þ

Fi g ure 9.2

α /2 = .0 25

α /2 = .0 2 5 z

-1.9 6 Re je ct H

0

1 .96 Rej e ct H0

0

c Similar to part a, with the rejection region in the lower tail of the D distribution. The null hypothesis H! will be rejected if D • • #Þ$$Þ d Similar to part b, with !Î # œ Þ ! ! & Þ The null hypothesis H! will be rejected if D ž # Þ & ) or D • • #Þ&) (which you can also write as lDl ž # Þ & )) Þ

9.2

a

The :-value for a right-tailed test is the area to the right of the observed test statistic D œ " Þ " & or : -value œ P(D ž " Þ " & Ñ œ " • Þ ) ( % * œ Þ " # & "

This is the shaded area in Figure *Þ$ .

185

Fig ure 9.3

p -valu e

z 0

1.1 5

b For a two-tailed test, the p-value is the probability of being as large or larger than the observed test statistic in either tail of the sampling distribution. As shown in Figure *Þ% , the p -value for D œ • #Þ() is : -value œ P(lDl ž # Þ ( ) Ñ œ # Ð Þ ! ! # ( Ñ œ Þ ! ! & %

Fig u re 9 .4

p -va lu e/2

p -va lu e /2

z -2 .7 8

c

0

2 .7 8

The :-value for a left-tailed test is the area to the left of the observed test statistic D œ • "Þ)" or : -value œ P(D • • "Þ)"ÑœÞ!$&"

9.3

Use the guidelines for statistical significance in Section *Þ$ . The smaller the p-value, the more evidence there is in favor of rejecting H! Þ For part a, p-value œÞ"#&" is not statistically significant; H! is not rejected. For part b, p -value œÞ!!&% is less than .!" and the results are highly significant; H! should be rejected. For part c, pvalue œÞ!$&" is between .!" and Þ!&. The results are significant at the & % level, but not at the " % level (P •Þ!&ÑÞ

9.4

In this exercise, the parameter of interest is ., the population mean. The objective of the experiment is to show that the mean exceeds #Þ$ .

186

a We want to prove the alternative hypothesis that . is, in fact, greater than #Þ$ . Hence, the alternative hypothesis is Ha : . ž # Þ $ and the null hypothesis is

b

H0 : . œ # Þ $ . _ The best estimator for . is the sample average B, and the test statistic is Dœ

B • .! 5/È8

_ which represents the distance (measured in units of standard deviations) from B to the hypothesized mean .. Hence, if this value is large in absolute value, one of two conclusions may be drawn. Either a very unlikely event has occurred, or the hypothesized mean is incorrect. Refer to part a. If ! œ .!& , the critical value of D that separates the rejection and non-rejection regions will be a value (denoted by D0 ) such that PÐD ž D0 Ñ œ ! œ .!& That is, D0 œ"Þ'%& (see Figure *Þ& ). Hence, H0 will be rejected if D ž " Þ ' % & .

Figure 9.5

α = .05

z 0

1.645 Reject H0

c The standard error of the mean is found using the sample standard deviation = to approximate the population standard deviation 5: SE œ

5 = Þ#* ¸ œ œÞ!%* È8 È8 È$&

d To conduct the test, calculate the value of the test statistic using the information contained in the sample. Note that the value of the true standard deviation, 5, is approximated using the sample standard deviation =. _ B • .! B • .! #Þ%•#Þ$ Dœ ¸ œ œ#Þ!% Þ!%* È È 5/ 8 =/ 8 The observed value of the test statistic, D œ # Þ ! % , falls in the rejection region and the null hypothesis is rejected. There is sufficient evidence to indicate that . ž # Þ $ .

187

9.5

a Since this is a right-tailed test, the p -value is the area under the standard normal distribution to the right of Dœ#Þ!%À : -value œ P(D ž # Þ ! % Ñ œ " • Þ * ( * $ œ Þ ! # ! ( b The p -value, Þ!#!( , is less than ! œ Þ ! & , and the null hypothesis is rejected at the & % level of significance. There is sufficient evidence to indicate that . ž#Þ$Þ c The conclusions reached using the critical value approach and the p -value approach are identical.

9.6

Refer to Exercise *Þ% , in which the rejection region was given as D ž " Þ ' % & where _ B • .! B•#Þ$ Dœ œ =/È8 .#*Î È$& _ Solving for B we obtain the critical value of B necessary for rejection of H0 . B•#Þ$ .#* ž"Þ'%& Ê B ž " Þ ' % & €#Þ$œ#Þ$) È È .#*/ $& $& b-c The probability of a Type II error is defined as " œ PÐaccept H0 when H0 is falseÑ Since the acceptance region is B Ÿ # Þ $ ) from part a, " can be rewritten as " œ PÐ B Ÿ # Þ $ ) when H0 is falseÑ œ PÐ B Ÿ # Þ $ ) when . ž#Þ$Ñ Several alternative values of . are given in this exercise. For . œ # Þ % , " œ PÐ B Ÿ # Þ $ ) when . œ # Þ % Ñ œ PÐD Ÿ œ PÐD Ÿ • .% " Ñ œ .$%!*

#Þ$)•#Þ% Ñ .#*/ È$&

For . œ # Þ $ , " œ PÐ B Ÿ # Þ $ ) when . œ # Þ $ Ñ œ PÐD Ÿ œ PÐD Ÿ " Þ ' $ Ñ œ .*%)%

#Þ$)•#Þ$ Ñ .#*/ È$&

For . œ # Þ & , " œ PÐ B Ÿ # Þ $ ) when . œ # Þ & Ñ œ PÐD Ÿ œ PÐD Ÿ • # %. & Ñ œ !. ! ( "

#Þ$)•#Þ& Ñ .#*/ È$&

For . œ # Þ ' , " œ PÐ B Ÿ # Þ $ ) when . œ # Þ ' Ñ œ PÐD Ÿ œ PÐD Ÿ • % %. * Ñ ¸ ! d

#Þ$)•#Þ' Ñ .#*/ È$&

The power curve is graphed using the values calculated above and is shown in Figure *Þ' .

188

Figure 9.6 1.0 0.9 0.8

Power

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0

2.3

9.7

2.4

2.5

µ

2.6

The hypotheses to be tested are H! À . œ # ) versus Ha À . Á # ) and the test statistic is Dœ

_ B • .! 5/È8

¸

B • .! =/È8

œ

#'Þ)•#) œ • "Þ)& 'Þ&Î È"!!

with p -value œ P(lDl ž " Þ ) & Ñ œ # Ð Þ ! $ # # Ñ œ Þ ! ' % % . To draw a conclusion from the p-value, u se the guidelines for statistical significance in Section *Þ$ . Since the p -value is greater than Þ!& , the null hypothesis should not be rejected. There is insufficient evidence to indicate that the mean is different from #) . (Some researchers might report these results as tending towards signficance.)

9.8

a If the airline is to determine whether or not the flight is unprofitable , they are interested in finding out whether or not . • ' ! (since a flight is profitable if . is at least '! ). Hence, the alternative hypothesis is Ha : . • ' ! and the null hypothesis is H0 : . œ ' ! . b Since only small values of B (and hence, negative values of D) would tend to disprove H0 in favor of Ha , this is a one-tailed test. c For this exercise, 8 œ " # ! , B œ &) and = œ " " . Hence, the test statistic is _ B • .! B • .! &)•'! Dœ ¸ œ œ • "Þ**# 5/È8 =/È8 ""Î È"#! The rejection region with ! œ .!& is determined by a critical value of D such thatPÐD • D0 Ñ œ .!&Þ This value is D0 œ •"Þ'%& and H! will be rejected if D • • "Þ'%& (compare the right-tailed rejection region in Figure *Þ& ). The observed value of D falls in the rejection region and H0 is rejected. The flight is unprofitable.

9.9

a

In order to make sure that the average weight was one pound, you would test H! À . œ " versus Ha À . Á "

b-c The test statistic is Dœ

_ B • .! 5/È8

¸

B • .! =/È8

189

œ

"Þ!"•" œÞ$$ Þ")Î È$&

with p -value œ P(lDl ž Þ $ $ Ñ œ # Ð Þ $ ( ! ( Ñ œ Þ ( % " % . Since the p -value is greater than Þ!& , the null hypothesis should not be rejected. The manager should report that there is insufficient evidence to indicate that the mean is different from"Þ

9.10

The theater chain claims that the average time is no more than $ minutes. To disprove this claim, you need to show that this average is more than $ minutes, and you should test H! À . œ $ versus Ha À . ž $ with the test statistic Dœ

_ B • .! 5/È8

¸

B • .! =/È8

œ

$Þ#&•$ œ$Þ&% !Þ&Î È&!

Since this is a one-tailed test, the rejection region with ! œ Þ ! " is set in the right tail of the D distribution as D ž DÞ!" œ # Þ $ $ (similar to Exercise *Þ" c). Since the observed value D œ $ Þ & % falls in the rejection region, H! is rejected. There is evidence that the average time is more than claimed by the theater chain.

9.11

a-b We want to test the null hypothesis that . is, in fact, )! % against the alternative that it is not: H! À . œ ) ! versus Ha À . Á ) ! Since the exercise does not specify . • ) ! or . ž ) ! , we are interested in a two directional alternative, . Á ) ! . c The test statistic is _ B • .! B • .! (*Þ(•)! Dœ ¸ œ œ • $Þ(& È È 5/ 8 =/ 8 Þ)Î È"!! The rejection region with ! œ .!& is determined by a critical value of D such that ! ! PÐD • • D0 Ñ € PÐD ž D0 Ñ œ € œ .!& # # This value is D0 œ " Þ * ' (see Figure *Þ# ). Hence, H0 will be rejected if D ž " Þ * ' or D • • "Þ*' . The observed value, D œ • $Þ(&, falls in the rejection region and H0 is rejected. There is sufficient evidence to refute the manufacturer's claim. The probability that we have made an incorrect decision is ! œ .!& .

9.12

The hypothesis to be tested is H! À . œ ( versus Ha À . • ( and the test statistic is Dœ

_ B • .! 5/È8

¸

B • .! =/È8

œ

'Þ(•( œ • Þ**# #Þ(Î È)!

The rejection region with ! œ .!& is D • • "Þ'%& (similar to Exercise *Þ) ). The observed value, D œ • .**# , does not fall in the rejection region and H0 is not rejected. The data do not provide sufficient evidence to indicate that . • ( .

9.13

a

The hypothesis to be tested is H! À . œ " " ! versus Ha À . • " " !

and the test statistic is Dœ

_ B • .! 5/È8

¸

B • .! =/È8

œ

"!(•""! œ • #Þ$" "$Î È"!!

with p -value œ P(D • • #Þ$"ÑœÞ!"!% . To draw a conclusion from the p-value, u se the guidelines for

190

statistical significance in Section *Þ$ . Since the p -value is between Þ!" and Þ!& , the test results are significant at the &% level, but not at the " % level. b If ! œ Þ ! & , H! can be rejected and you can conclude that the average score improvement is less than claimed. This would be the most beneficial way for the competitor to state these conclusions. c If you worked for the Princeton Review, it would be more beneficial to conclude that there was insufficient evidence at the 1% level to conclude that the average score improvement is less than claimed.

9.14

a-b The hypothesis of interest is one-tailed: H0 : .1 • .2 œ ! versus Ha : .1 • .2 ž ! c

The test statistic, calculated under the assumption that .1 • .2 œ ! , is Dœ

(B" • B# ) • (.1 • .2 ) Ë

521 81

€

522 82

with 521 and 522 known, or estimated by =21 and =22 , respectively. For this exercise, D¸

(B" • B# ) • ! Ë

=21 81

€

=22

œ

82

""Þ'•*Þ(

#(Þ* $)Þ% Ê € )! )!

œ#Þ!*

a value which lies slightly more than two standard deviations from the hypothesized difference of zero. This would be a somewhat unlikely observation, if H! is true. d The p -value for this one-tailed test is p-value œ P(D ž # Þ ! * Ñ œ " • Þ * ) " ( œ Þ ! " ) $ Since the p -value is not less than ! œ Þ ! " , the null hypothesis cannot be rejected at the " % level. There is insufficient evidence to conclude that ." • .# ž ! Þ e Using the critical value approach, the rejection region, with ! œ .!" , is D ž # Þ $ $ (see Exercise *Þ" a).Since the observed value of D does not fall in the rejection region, H0 is not rejected. There is insufficient evidence to indicate that ." • .# ž ! , or .1 ž .2 .

9.15

The hypothesis of interest is one-tailed: H0 : .1 • .2 œ ! versus Ha : .1 • .2 • ! The test statistic, calculated under the assumption that .1 • .2 œ ! , is D¸

(B" • B# ) • ! Ë

=21 81

€

=22

œ

82

"Þ#%•"Þ$"

Þ!&' Þ!&% Ê € $' %&

œ • "Þ$$

withthe unknown 521 and 522 estimated by s 21 and s 22 , respectively. The student can use one of two methods for decision making. p-value approach: Calculate p -value œ P(D • • "Þ$$ÑœÞ!*")Þ Since this p -value is greater than Þ!& , the null hypothesis is not rejected. There is insufficient evidence to indicate that the mean for population " is smaller than the mean for population # . Critical value approach: The rejection region, with ! œ .!& , is D • • "Þ'%&Þ Since the observed value of D does not fall in the rejection region, H0 is not rejected. There is insufficient evidence to indicate that the mean for population " is smaller than the mean for population # .

191

9.16

The probability that you are making an incorrect decision is influenced by the fact that if .1 • .2 œ ! , it is just as likely that B" • B# will be positive as that it will be negative. Hence, a two-tailed rejection region must be used. Choosing a one-tailed region after determining the sign of B" • B# simply tells us which of the two pieces of the rejection region is being used. Hence, ! œ PÐreject H0 when H0 trueÑ œ PÐD ž " Þ ' % & or D • • "Þ'%& when H0 true Ñ œ !1 € !2 œ .! & € Þ ! & œ Þ " ! which is twice what the experimenter thinks it is. Hence, one cannot choose the rejection region after the test is performed.

9.17

a

The hypothesis of interest is one-tailed: H0 : .1 • .2 œ ! versus Ha : .1 • .2 ž !

b

The test statistic, calculated under the assumption that .1 • .2 œ ! , is D¸

(B" • B# ) • ! Ë

=21 81

€

=22

œ

82

'Þ*•&Þ) Ð#Þ*Ñ# Ð"Þ#Ñ# Ë € $& $&

œ#Þ!(%

The rejection region, with ! œ .!& , is D ž " Þ ' % & and H0 is rejected. There is evidence to indicate that .1 • .2 ž ! , or .1 ž .2 . That is, there is reason to believe that Vitamin C reduces the mean time to recover.

9.18

a

The hypothesis of interest is one-tailed: H0 : .1 • .2 œ ! versus Ha : .1 • .2 ž !

The test statistic, calculated under the assumption that .1 • .2 œ ! , is D¸

(B" • B# ) • ! Ë

=21 81

€

=22

œ

82

($•'$ Ë

Ð#&Ñ# %!!

€

œ&Þ$$

Ð#)Ñ# %!!

The rejection region, with ! œ .!" , is D ž # Þ $ $ and H0 is rejected. There is evidence to indicate that .1 • .2 ž ! , or .1 ž .2 .The average per-capita beef consumption has decreased in the last ten years. (Alternatively, the p -value for this test is the area to the right of D œ & Þ $ $ which is very close to zero and less than ! œÞ!"ÞÑ b For the difference .1 • .2 in the population means this year and ten years ago, the ** % lower confidence bound uses DÞ!" œ#Þ$$ and is calculated as =21 =22 _ _ #& # #) # Ë (B1 • B2 ) •#Þ$$ € œ (( $ • ' $ ) •#Þ$$ Ë € 81 82 %!! %!! "!•%Þ$( or (.1 • .2 Ñ ž & Þ ' $ Since the difference in the means is positive, you can again conclude that there has been a decrease in the average per-capita beef consumption over the last ten years. In addition, it is likely that the average consumption has decreased by more than &Þ'$ pounds per year.

9.19

a

The hypothesis of interest is two-tailed: H0 : .1 • .2 œ ! versus Ha : .1 • .2 Á !

192

The test statistic, calculated under the assumption that .1 • .2 œ ! , is D¸

(B" • B# ) • ! Ë

=21 81

€

=22

œ

82

$%Þ"•$' Ë

Ð&Þ*Ñ# "!!

€

œ • #Þ#'

Ð'Þ!Ñ# "!!

with p -value œ P(lDl ž # Þ # ' Ñ œ # Ð Þ ! " " * Ñ œ Þ ! # $ ) . Since the p -value is less than Þ!& , the null hypothesis is rejected. There is evidence to indicate a difference in the mean lead levels for the two sections of the city. b From Section )Þ' , the *& % confidence interval for .1 • .2 is approximately =21 =22 _ _ Ë (B1 • B2 ) „"Þ*' € 81 82 ($ % Þ " • $ ' ) „"Þ*' Ë •"Þ*„"Þ'&

Ð&Þ*Ñ# Ð'Þ!Ñ# € "!! "!! or • $ Þ & & • (.1 • .2 ) • •Þ#&

c Since the value .1 • .2 œ & or .1 • .2 œ • & is not in the confidence interval in part b, it is not likely that the difference will be more than & ppm, and hence the statistical significance of the difference is not of practical importance to the engineers.

9.20

The hypothesis of interest is two-tailed:: H0 : .1 • .2 œ ! versus Ha : .1 • .2 Á ! and the test statistic, calculated under the assumption that .1 • .2 œ ! , is D¸

(B" • B# ) • ! Ë

=21 81

€

=22

œ

82

$&ß&&%•$$ß$%) ###& # #$(& # € Ë &! &!

œ%Þ(*

The rejection region, with ! œ .!& , is ± D ± ž"Þ*' and H0 is rejected. There is evidence to indicate a difference in the means for the graduates in education and the social sciences. b The conclusions are the same.

9.21

a Most people would have no preconceived idea about which of the two hotels would have higher average room rates, and a two-tailed hypothesis would be appropriate: H0 : .1 • .2 œ ! versus Ha : .1 • .2 Á ! b

The test statistic is D¸

(B" • B# ) • ! Ë

=21 81

€

=22

œ

82

"#!•*& Ë

"(Þ& # &!

€

œ)Þ((

"! # &!

The rejection region, with ! œ .!" , is lDl ž # Þ & ) and H0 is rejected. There is evidence to indicate that there is a difference in the average room rates for the Marriott and the Radisson hotels. c The p -value for this two-tailed test is p-value œ P(D ž ) Þ ( ( Ñ € PÐD • • )Þ((ѸÞ!!!! Since the p -value is less than ! œ Þ ! " , the null hypothesis cannot be rejected at the " % level. There is sufficient evidence to conclude that ." • .# Á ! Þ

193

9.22

a

The two-tailed hypothesis is H0 : .1 • .2 œ ! versus Ha : .1 • .2 Á !

and test statistic is D¸

(B" • B# ) • ! Ë

=21 81

€

=22

œ

82

""!•*& "'Þ& # "! # € Ë &! &!

œ&Þ&!

The rejection region, with ! œ .!& , is lDl ž " Þ * ' and H0 is rejected. There is evidence to indicate that there is a difference in the average room rates for the Wyndham and the Radisson hotels. b The *& % confidence interval for .1 • .2 is approximately =21 =22 _ _ (B1 • B2 ) „"Þ*' Ë € 81 82 (" " ! • * & ) „"Þ*' Ë

"'Þ& # "! # € &! &! "&„&Þ$%) or *Þ'&# • (.1 • .2 ) • #!Þ$%)

Since the value .1 • .2 œ ! is not in the confidence interval, it is not likely that the two means are the same. This confirms the conclusion in part a.

9.23

a

The hypothesis of interest is two-tailed: H0 : .1 • .2 œ ! versus Ha : .1 • .2 Á !

and the test statistic is D¸

(B" • B# ) • ! Ë

=21 81

€

=22

œ

82

Þ*%•#Þ) Ë

"Þ# # $'

€

œ • $Þ")

#Þ) # #'

with p -value œ P(lDl ž $ Þ " ) Ñ œ # Ð Þ ! ! ! ( Ñ œ Þ ! ! " % . Since the p -value is less than Þ!& , the null hypothesis is rejected. There is evidence to indicate a difference in the meanconcentrations for these two types of sites. b The *& % confidence interval for .1 • .2 is approximately =21 =22 _ _ Ë (B1 • B2 ) „"Þ*' € 81 82 (.* % • # Þ ) ) „"Þ*' Ë •"Þ)'„"Þ"&

"Þ# # #Þ) # € $' #' or • $ Þ ! " • (.1 • .2 ) • •Þ("

Since the value .1 • .2 œ ! does not fall in the interval in part b, it is not likely that .1 œ .2 Þ There is evidence to indicate that the means are different, confirming the conclusion in part a.

9.24

a

The hypothesis of interest concerns the binomial parameter : and is one-tailed: H0 : : œ .$ versus

b

Ha : : • .$

The rejection region is one-tailed, with ! œ .!& or D • • "Þ'%&Þ

194

c

It is given that B œ # ( * and 8 œ " ! ! ! , so that s :œ Dœ

s : • :0

: 0 ;0 Ê 8

B #(* œ œ .#(* . The test statistic is then 8 "!!!

.# ( * • .$

œ

.$ (.( ) Ê "!!!

œ • "Þ%%*

Since the observed value does not fall in the rejection region, H0 is not rejected. We cannot conclude that :•Þ$.

9.25

a

The hypothesis of interest is two-tailed: H0 : : œ .% versus

b-c It is given that B œ & # * and 8 œ " % ! ! , so that s :œ Dœ

s : • :0

: 0 ;0 Ê 8

Ha : : Á .%

B &#* œ œ .$() . The test statistic is 8 "%!!

.$ ( ) • .%

œ

.% (.' ) Ê "%!!

œ • "Þ')

with p -value œ P(lDl ž " Þ ' ) Ñ œ # Ð Þ ! % ' & Ñ œ Þ ! * $ Þ Since the p -value is not less than ! œ Þ ! " , the null hypothesis is not rejected. There is insufficient evidence to indicate that : differs from .% .

9.26

The hypothesis of interest is one-tailed: H0 : : œ .& versus sœ With B œ ( # and 8 œ " # ! , so that :

Ha : : ž Þ &

B (# œ œ .' , the test statistic is 8 "#! Dœ

s : • :0

œ

: 0 ;0 Ê 8

.' • .&

.& (.& ) Ê "#!

œ#Þ"*

The student may use one of two approaches: Critical value approach: The rejection region is one-tailed, D ž " Þ ' % & with ! œ .!& or D ž # Þ $ $ with ! œÞ!"Þ Hence, H0 is rejected at the & % level, but not at the " % level. At the & % significance level, we conclude that : ž .&. p-value approach: Calculate p -value œ P(D ž # Þ " * Ñ œ " • Þ * ) & ( œ Þ ! " % $ Þ Since this p -value is between Þ!" and Þ!&, H0 is rejected at the & % level, but not at the " % level. At the & % significance level, we conclude that : ž .&.

9.27

a

b

The two sets of hypotheses both involve a different binomial parameter : : H0 : : œ .' versus

Ha : : Á Þ ' (part c)

H0 : : œ .& versus

Ha : : • Þ & (part b)

For the second test in part a, B œ $ & and 8 œ ( &, so that s :œ Dœ

s : • :0

: 0 ;0 Ê 8

œ

.%''(• .& .& (.& ) Ê (&

B $& œ œ .%''( , the test statistic is 8 (& œ • Þ&)

Since no value of ! is specified in advance, we calculate p -value œ P(D • • Þ&)ÑœÞ#)"!Þ Since this p -value is greater than Þ"! , the null hypothesis is not rejected. There is insufficient evidence to contradict the claim.

195

c

For the first test in part a, B œ % * and 8 œ ( &, so that s :œ Dœ

s : • :0

: 0 ;0 Ê 8

œ

B %* œ œ .'&$$ , the test statistic is 8 (&

.'&$$• .' .' (.% ) Ê (&

œÞ*%

with p -value œ P(lDl ž Þ * % Ñ œ # Ð Þ " ( $ ' Ñ œ Þ $ % ( # Þ Since this p -value is greater than Þ"! , the null hypothesis is not rejected. There is insufficient evidence to contradict the claim.

9.28

a

The hypothesis of interest is two-tailed: H0 : : œ .(& versus

b

sœ With B œ & ) and 8 œ " ! ! , so that : Dœ

Ha : : Á Þ ( &

B &) œ œ .&) , the test statistic is 8 "!!

s : • :0

: 0 ;0 Ê 8

œ

.& ) • (. &

.(& (.#& ) Ê "!!

œ • $Þ*$

with p -value œ P(lDl ž $ Þ * $ Ñ • # Ð Þ ! ! ! # Ñ œ Þ ! ! ! % or : -value ¸ ! Þ Since this p -value is less than Þ!" , H! is rejected at the " % level of signicance and the results are declared highly significant. There is evidence that the proportion of red flowered plants is not Þ(& .

9.29

a-b Since the survival rate without screening is : œ # Î $ , the survival rate with an effectiveprogram may be greater than #Î$ . Hence, the hypothesis to be tested is H0 : : œ # Î $ versus c

With s :œ

Ha : : ž # Î $

B "'% œ œ .)# , the test statistic is 8 #!! Dœ

s : • :0

: 0 ;0 Ê 8

œ

.) # • # Î $ Ð#Î$ÑÐ"Î$Ñ Ê #!!

œ%Þ'

The rejection region is one-tailed, with ! œ .!& or D ž " Þ ' % & and H0 is rejected. The screening program seems to increase the survival rate. d For the one-tailed test, p-value œ PÐD ž % Þ ' Ñ • " • *. * * ) œ .!!!# .!!!# .

That is, H0 can be rejected for any value of !

9.30

a

The hypothesis of interest is one-tailed: H0 : : œ ." versus

With s :œ

Ha : : ž Þ "

B #& œ œ .#& , the test statistic is 8 "!! Dœ

s : • :0

: 0 ;0 Ê 8

œ

.# & •. "

." (.* ) Ê "!!

œ&

with p -value œ P(D ž & Ñ • " • Þ * * * ) œ Þ ! ! ! # Þ Since this p -value is less than Þ!& , H! is rejected at the & % level of signicanceÞ There is evidence that the proportion of infested fields is larger than expected. b The most obvious reason for this unusually high proportion of infested fields is that there is contagion at work; that is, the 8 œ " ! ! fields may not be independent, but may be contaminating one another.

196

9.31

The hypothesis of interest is H0 : : œ .%& versus With s :œ

Ha : : Á Þ % &

B $# œ œ .% , the test statistic is 8 )! Dœ

s : • :0

.% ! • %. &

œ

: 0 ;0 Ê 8

.%& (.&& ) Ê )!

œ • Þ*!

The rejection region is two-tailed with ! œ .!" , or ± D ± ž#Þ&) and H0 is not rejected. There is insufficient evidence to dispute the newspaper's claim.

9.32

a

The hypothesis of interest is H0 : : œ .#& versus

With s :œ

Ha : : Á Þ # &

B ## œ œ .#(& , the test statistic is 8 )! Dœ

s : • :0

: 0 ;0 Ê 8

œ

.# ( & • .#&

.#& (.(& ) Ê )!

œÞ&#

The rejection region is two-tailed with ! œ .!& , or ± D ± ž"Þ*' and H0 is not rejected. There is insufficient evidence to indicate that the claim is incorrect. b The hypothesis of interest is H0 : : œ .#% versus With s :œ

Ha : : Á Þ # %

B "& œ œ .")(& , the test statistic is 8 )! Dœ

s : • :0

: 0 ;0 Ê 8

œ

.")(&• .#% .#% (.(' ) Ê )!

œ • "Þ"!

The rejection region is two-tailed with ! œ .!& , or ± D ± ž"Þ*' and H0 is not rejected. There is insufficient evidence to indicate that the claim is incorrect. c Unless the experimenter had some preconceived idea that the proportion might be greater or less than claimed, there would be no reason to run a one-tailed test.

9.33

The hypothesis of interest is H0 : : œ ."' versus with s :œ

Ha : : Á Þ " '

B "%# œ œÞ"%# ,and the test statistic is 8 "!!! Dœ

s : • :0

: 0 ;0 Ê 8

œ

." % # • ."'

."' (.)% ) Ê "!!!

œ • "Þ&&

Since no value of ! is specified in advance, we calculate p -value œ P(lDl ž " Þ & & Ñ œ # Ð Þ ! ' ! ' Ñ œ Þ " # " # Þ Since this p-value is greater than Þ"! , the null hypothesis is not rejected. There is insufficient evidence to indicate that the percentage of male elementary-school teachers in California is different from the national percentage.

197

9.34

a

Since it is necessary to detect either : 1 ž :2 or : 1 • : 2 , a two-tailed test is necessary: H0 : : 1 • : 2 œ !

b

Ha : : 1 • : 2 Á !

versus

The standard error of s : " • :s# is Ê

: " ;" : # ;# € 8" 8#

In order to evaluate the standard error, estimates for : 1 and : 2 must be obtained, using the assumption that : 1 • : 2 œ ! . Because we are assuming that : 1 œ : 2 , the best estimate for this common value will be B1 € B2 (%€)" œ œ .&&% 81 € 82 "%!€"%!

s :œ and the estimated standard error is

" " # :;Œ € • œ ËÞ&&%ÐÞ%%'ÑŒ • œÞ!&*% Ëss 8" 8# "%! c

Calculate s :1 œ

(% )" œ .&#* and s :2 œ œ .&(*Þ The test statistic, based on the sample data will be "%! "%! Dœ

s :1 • s : 2 • (: 1 • : 2 ) : 1 ;1 : 2 ;2 € Ê 81 82

¸

s : 1 • :s2

:;Š Ëss

" " € ‹ 81 82

œ

.& # * • .&(* œ • .)% Þ!&*%

This is a likely observation if H! is true, since it lies less than one standard deviation below : " • : # œ ! Þ d Calculatethe two tailed p-value œ P(lDl ž Þ ) % Ñ œ # Ð Þ # ! ! & Ñ œ Þ % ! " ! Þ Since this p -value is greater than Þ!" , H! is not rejected. There is no evidence of a difference in the two population proportions. e The rejection region with ! œ .!" is lDl ž # Þ & ) and H0 is not rejected. There is no evidence of a difference in the two population proportions.

9.35

a-b If : 1 cannot be larger than : 2 , the only alternative to H0 : : 1 • : 2 œ ! is that : 1 • : 2 , and the one-tailed alternative is Ha : : 1 • : 2 • ! . c The rejection region, with ! œ Þ ! & , is D • • "Þ'%& and the observed value of the test statistic is D œ • .)% . The null hypothesis is not rejected. There is no evidence of to indicate that : " is smaller than : # Þ

9.36

The hypothesis of interest is one-tailed: H0 : : 1 • : 2 œ !

versus

Ha : : 1 • : 2 • !

"$# "() B1 € B2 "$#€"() œ .%(" , s :2 œ œ .&!* , and s :œ œ œ .%*# #)! $&! 81 € 82 #)!€$&! The test statistic is then Calculate s :1 œ

Dœ

s : 1 • :s2

:;Š Ëss

" " € ‹ 81 82

œ

.% ( " • .&!*

È.%*# (.&!) )(" /# ) ! € " /$ & ! )

œ • .*&

Since no value of ! is specified in advance, we calculate p -value œ P(D • • Þ*&ÑœÞ"(""Þ Since this p -value is greater than Þ"! , the null hypothesis is not rejected. There is insufficient evidence to indicate that : 1 • : 2 .

9.37

a

The hypothesis of interest is H0 : : 1 • : 2 œ !

s2 œ .'! , and : sœ Calculate s : 1 œ .$' , :

versus

81 s:1 € 82 s:2 81 € 82 198

œ

Ha : : 1 • : 2 • ! ")€$! œ .%)Þ The test statistic is then &!€&!

Dœ

s : 1 • :s2

:;Š Ëss

" " € ‹ 81 82

œ

.$ ' • '. ! œ • #Þ%! È.%) (.&# )(" /& ! € " /& ! )

The rejection region with ! œ .!& is D • • "Þ'%& and H0 is rejected. There is evidence of a difference in the proportion of survivors for the two groups. b From Section )Þ( , the approximate *& % confidence interval is (s :1 • s : 2 ) „"Þ*' Ë

(.$ ' • .' ! ) „"Þ*' Ê

a

81

€

s : 2 s;2 82

.$' (.'% ) .'! (.%! ) € &! &! • Þ % $ • ( :1 • : 2 ) • • Þ!&

• Þ # % „ ."* or

9.38

s : 1 s;1

The hypothesis of interest is H0 : : 1 • : 2 œ !

versus

Ha : : 1 • : 2 Á !

"#$ "%& B1 € B2 "#$€"%& œ .#)! , s :2 œ œ .#&* , and s :œ œ œ .#') %%! &'! 81 € 82 %%!€&'! The test statistic is then Calculate s :1 œ

Dœ

s : 1 • :s2

" " :;Š € ‹ Ëss 81 82

.# ) ! • .#&*

È.#') (.($# )(" /% % ! € " /& ' ! )

œ

œ!Þ(%

The rejection region with ! œ .!" is lDl ž # Þ & ) and H0 is not rejected. There is no evidence of a difference in the proportionof frequent moviegoers in the two demographic groups. b A difference in the proportions might mean that the advertisers would choose different products to advertise before this movie.

9.39

The hypothesis of interest is H0 : : 1 • : 2 œ !

versus

Ha : : 1 • : 2 Á !

"# ) B1 € B2 "#€) œ .#"% , s :2 œ œ .#& , and s :œ œ œ .##( &' $# 81 € 82 &'€$# The test statistic is then Calculate s :1 œ

Dœ

s : 1 • :s2

:;Š Ëss

" " € ‹ 81 82

.# " % • .#& œ • Þ$* È.##( (.(($ )(" /& ' € " /$ # )

œ

The rejection region with ! œ .!& is lDl ž " Þ * ' and H0 isnot rejected. There is insufficient evidence to indicate a difference in the proportion of red M&Ms for the plain and peanut varieties. These results match the conclusions of Exercise )Þ%( .

9.40

a

The hypothesis of interest is H0 : : 1 • : 2 œ !

versus

199

Ha : : 1 • : 2 Á !

s2 œ .&' , and : sœ Calculate s : 1 œ .'' , :

81 s:1 € 82 s:2 81 € 82

œ

#$'ÐÞ''Ñ€#(#ÐÞ&'Ñ œ .'!'Þ The test statistic is #$'€#(#

then Dœ

s : 1 • :s2

" " :;Š € ‹ Ëss 81 82

.' ' • &. '

È.'!' (.$*% )(" /# $ ' € " /# ( # )

œ

œ#Þ$!

with p -value œ P(lDl ž # Þ $ ! Ñ œ # Ð Þ ! " ! ( Ñ œ Þ ! # " % Þ Since the p -value is between Þ!" and Þ!& , the results are reported significant at the & % level of significance, but not at the " % level. There is evidence of a difference in the proportion of boys and girls who "surf the net". b The hypothesis of interest is H0 : : 1 • : 2 œ ! s2 œ .%' , and : sœ Calculate s : 1 œ .&( , :

versus

81 s:1 € 82 s:2 81 € 82

œ

Ha : : 1 • : 2 Á ! #$'ÐÞ&(Ñ€#(#ÐÞ%'Ñ œ .&""Þ The test statistic is #$'€#(#

then Dœ

s : 1 • :s2

:;Š Ëss

" " € ‹ 81 82

œ

.& ( • %. ' œ#Þ%( È.&"" (.%)* )(" /# $ ' € " /# ( # )

with p -value œ P(lDl ž # Þ % ( Ñ œ # Ð Þ ! ! ' ) Ñ œ Þ ! " $ ' Þ Since the p -value is between Þ!" and Þ!& , the results are reported significant at the & % level of significance, but not at the " % level. There is evidence of a difference in the proportion of boys and girls who think technology makes a positive difference in their lives. c Since the same teenagers were asked both questions, the two tests are not independent. If you happen to have a sample that contains only a small number of teenagers who surf the net, the proportions answering yes to the second question may also be small, and vice versa.

9.41

If there is no statistically significant difference between the percentage of teenagers using computers for lowerincome versus wealthier families, this would indicate that the cost of the computer is not a factor that affects this percentage. Since many computers are used in middle school or high school classrooms, this result would indicate that classroom environments are playing a significant role in educating students in computer skills.

9.42

a Since the two treatments were randomly assigned, the randomization procedure can be implemented as each patient becomes available for treatment. Choose a random number between ! and * for each patient. If the patient receives a number between ! and % , the assigned drug is aspirin. If the patient receives a number between & and * , the assigned drug is clopidogrel. b Assume that 8" œ * * # & and 8# œ*#'!Þ Calculate s : 1 œ .!&$ , s : 2 œ .!&) , and s :œ

81 s:1 € 82 s:2 81 € 82

œ

**#&ÐÞ!&$Ñ€*#'!ÐÞ!&)Ñ œ .!&&Þ "*")&

The test statistic is then Dœ

s : 1 • :s2

" " :;Š € ‹ Ëss 81 82

œ

.! & $ • .!&)

È.!&& (.*%& )(" /* * # & € " /* # ' ! )

œ • "Þ&#

with p -value œ P(lDl ž " Þ & # Ñ œ # Ð Þ ! ' % $ Ñ œ Þ " # ) ' Þ Since the p -value is greater than Þ"! , the results are not statistically significant. There is insufficient evidence to indicate a difference in the proportions for the two treatment groups.

200

9.43

The hypothesis of interest is H0 : : 1 • : 2 œ ! Calculate s :1 œ

versus

Ha : : 1 • : 2 ž !

*$ ""* B1 € B2 *$€""* œÞ('*ß :s2 œ œ Þ & * ) , and s :œ œ œ .''#&Þ The test "#" "** 81 € 82 "#"€"**

statistic is then Dœ

s : 1 • :s2

" " :;Š € ‹ Ëss 81 82

œ

.( ' * • .&*)

È.''#& (.$$(& )(" /" # " € " /" * * )

œ$Þ"%

with p -value œ P(D ž $ Þ " % Ñ œ " • Þ * * * # œ Þ ! ! ! ) Þ Since the p -value is less than Þ!" , the results are reported as highly significant at the " % level of significanceÞ There is evidence to confirm the researcher's conclusion.

9.44

The probability of a Type I error is ! œ PÐreject H! when H! is true) and the probability of a Type II error is " œ PÐaccept H! when H! is false). a " increases. b In order to decrease both ! and ", you must increase the sample size.

9.45

See Section *Þ$ of the text.

9.46

The following conditions must hold in order that the D statistic be appropriate: 1 B is a normally distributed random variable (the Central Limit Theorem insures this as long as 8 $ ! and the sample is random). 2 5 is a known quantity, or 8 is large, so that a good approximation for 5_B can be obtained from the sample observations.

9.47

The power of the test is " • " œ PÐreject H! when H! is false). As . gets farther from .! ß the power of the test increases.

9.48

The hypothesis to be tested is H0 : . œ & Þ ( versus Ha : . • & Þ ( and the test statistic is Dœ

B•.

5/È8

¸

B•.

=/È8

œ

$Þ(•&Þ( œ • #&Þ#*) .&/ È%!

The rejection region with ! œ .!& is D • • "Þ'%& . The observed value, D œ • #&Þ#*) , falls in the rejection region and H0 is rejected. We conclude that the average pH for rains is more acidic than for pure water.

9.49

The objective of this experiment is to make a decision about the binomial parameter : , which is the probability that a customer prefers the first color. Hence, the null hypothesis will be that a customer has no preference for the first color, and the alternative will be that he does have a preference. If the null hypothesis is true, then H0 : : œ P[customer prefers the first color] œ " Î $ If the customer actually has a preference for the first color, then Ha : : ž " Î $ a

The test statistic is calculated with s :œ

%!! œ .% as "!!!

201

Dœ

s : • :0

.% • " Î $

œ

: 0 ;0 Ê 8

("Î$ )(#Î$ ) Ê "!!!

œ%Þ%(

and the p -value is p-value œ PÐD ž % Þ % ( Ñ • " • *. * * ) œ .!!!# since PÐD ž % Þ % ( Ñ is surely less than PÐD ž $ Þ % * Ñ, the largest value in Table $ . b Since ! œ .!& is larger than the p -value, which is less than .!!!# , H0 can be rejected. We conclude that customers have a preference forthe first color.

9.50

a

The hypothesis of interest is H0 : : 1 • : 2 œ !

We have s : 1 œ Þ # ! ß :s2 œ Þ # ' , and s :œ

versus

Ha : : 1 • : 2 Á !

81 s : " € 82 s:# &!!ÐÞ#!Ñ€&!!ÐÞ#'Ñ œ œ .#$Þ The test statistic is 81 € 82 &!!€&!!

then Dœ

s : 1 • :s2

:;Š Ëss

" " € ‹ 81 82

œ

.# ! • #. ' œ • #Þ#& È.#$ (.(( )(" /& ! ! € " /& ! ! )

The rejection region with ! œ Þ ! " is lDl ž # Þ & ) and H! is not rejected. There is insufficient evidence to indicate a difference in the proportion of men and women who think that space should remain commercial free. b Answers will vary.

9.51

a-b Since it is necessary to prove that the average pH level is less than (Þ& , the hypothesis to be tested is onetailed: H0 : . œ ( Þ & versus Ha : . • ( Þ & d

The test statistic is Dœ

B•.

5/È8

¸

B•.

=/ È8

œ

• .# œ • &Þ%(( .#/ È$!

and the rejection region with ! œ .!& is D • • "Þ'%& . The observed value, D œ • &Þ%(( , falls in the rejection region and H0 is rejected. We conclude that the average pH level is less than (Þ& .

9.52

a

The hypothesis of interest is H0 : : œ .& versus

with s :œ

Ha : : • Þ &

B ' œ œÞ### ,and the test statistic is 8 #( Dœ

s : • :0

: 0 ;0 Ê 8

œ

.# # # • .& .& (.& ) Ê #(

œ • #Þ)*

Since no value of ! is specified in advance, we calculate p -value œ P(D • • # Þ ) * Ñ œ Þ & • Þ % * ) " œ Þ ! ! " * Þ Since this p -value isless than Þ!" , you can reject H! at the " % level (highly significant) and reject the DA's claim of &! % or greater. B %&& b If you take B œ % & & and 8 œ & ! $ with s :œ œ œ Þ * ! & , the *& % confidence interval for : is 8 &!$ approximately

202

s : „ " Þ * 'Ë

ss :; .*!& (.!*& ) œ .*!&„"Þ*' Ê œ .* ! & „ .!#' 8 &!$

or .) ( * • : • .*$" . c Even with the conservative value of B in part b, you can see that all the possible values for : are greater than : œ Þ & . You cannot conclude that the DA's claim of &! % or greater is wrong—in fact, it appears that the DA is correct!

9.53

a-b Since there is no prior knowledge as to which mean should be larger, the hypothesis of interest is two-tailed H0 : .1 • .2 œ ! versus Ha : .1 • .2 Á ! c

The test statistic is D¸

(B" • B# ) • ! Ë

=21 81

€

=22

œ

82

#*)!•$#!& ""%! # *'$ # € Ë %! %!

œ • Þ*&%

The rejection region, with ! œ .!& , is two-tailed or ± D ± ž"Þ*' . The null hypothesis is not rejected. There is insufficient evidence to indicated a difference in the two means.

9.54

The experimenter has two binomial populations where : 1 is the proportion judging the automobile more expensive for group A and : 2 is the proportion judging the automobile more expensive for group B. The hypothesis to be tested is H0 : : 1 • : 2 œ ! Calculate s :1 œ

versus

Ha : : 1 • : 2 ž !

$( #$ B1 € B2 $(€#$ œ Þ ( % ß :s2 œ œ Þ % ' , and s :œ œ œ .'Þ The test statistic is then &! &! 81 € 82 &!€&! Dœ

s : 1 • :s2

:;Š Ëss

.( % • %. ' œ#Þ)&) È.' (.% )(" /& ! € " /& ! )

œ

" " € ‹ 81 82

The rejection region with ! œ .!& is D ž " Þ ' % & and H0 is rejected. There is sufficient evidence to indicate that using a female model influences the perceived expensiveness of an automobile.

9.55

Let : 1 be the proportion of defectives produced by machine A and : 2 be the proportion of defectives produced by machine B. The hypothesis to be tested is H0 : : 1 • : 2 œ ! Calculate s :1 œ

versus

Ha : : 1 • : 2 Á !

"' ) B1 € B2 "'€) œ Þ ! ) ß :s2 œ œ Þ ! % , and s :œ œ œ .!'Þ The test statistic is #!! #!! 81 € 82 #!!€#!!

then Dœ

s : 1 • :s2

" " :;Š € ‹ Ëss 81 82

œ

.! ) • !. %

È.!' (.*% )(" /# ! ! € " /# ! ! )

œ"Þ')%

The rejection region with ! œ .!& is lDl ž " Þ * ' and H0 is not rejected. There is insufficient evidence to indicate that the machines are performing differently in terms of the percentage of defectives being produced.

9.56

a b

The hypothesis to be tested is H0 : . œ $ & versus Ha : . • $ & . The rejection region with ! œ .01 is D • • #Þ$$ .

203

c

The test statistic is Dœ

B•.

5/È8

¸

B•.

=/È8

œ

$"Þ(&•$& œ • 'Þ"* "!Þ& /È%!!

The observed value, D œ • 'Þ"* , falls in the rejection region and H0 is rejected. There is sufficient evidence to indicate that the mean is less than $& kg/m# .

9.57

a

The hypothesis to be tested is H0 : : 1 • : 2 œ !

Calculate s :1 œ

versus

Ha : : 1 • : 2 ž !

%' $% B1 € B2 %'€$% œ Þ # $ ß :s2 œ œ Þ " ( , and s :œ œ œ .#Þ The test statistic is #!! #!! 81 € 82 #!!€#!!

then Dœ

s : 1 • :s2

:;Š Ëss

" " € ‹ 81 82

.# $ • ". ( œ"Þ& È.# (.) )(" /# ! ! € " /# ! ! )

œ

and the p -value is PÐD " Þ & Ñ œ " • *. $ $ # œ .!'') . b Sincethe observed p -value, .!'') , is greater than ! œ .!& , H0 cannot be rejected. There is insufficient evidence to support the researcher's belief.

9.58

Refer to Figure *Þ( , which represents the two probability distributions, one assuming that : 1 • : 2 œ ! and one assuming that : 1 • : 2 œÞ"Þ Figure 9.7

β = .20

A

α = .05

.1

0

The right curve is the true distribution of the random variable s : 1 • :s2 and consequently any probabilities that we wish to calculate concerning the random variable must be calculated as areas under the curve to the right. The objective of this exercise is to find a common sample size so that ! œ P[reject H0 when H0 true] œ . !& and " œ P[accept H0 when H0 false] Ÿ . #! .For ! œ .!& consider the critical value of s : 1 • :s2 that separates the s s rejection and acceptance regions. This value will be denoted by (: 1 • : 2 )G . Recall that the random variable D œ (B • .)/ 5 measures the distance from a particular value B to the mean (in units of standard deviation). Since the D-value corresponding to (s :1 • s : 2 )G is D œ " Þ ' % & , we have "Þ'%& œ

(s :1 • s : 2 )G • !

Ê

204

: 1 ;1 : 2 ;2 € 81 82

or (s :1 • s : 2 )G œ"Þ'%& Ê

: 1 ;1 : 2 ;2 € . 81 82

Now " œ P[accept H0 when : 1 • : 2 œ ." ]which is the area under the right hand curve to the left of (s :1 • s : 2 )G . Since it is required that " œ .#! we must find the D-value corresponding to A œ .# , which is D œ • .)% (see Table $ ). Then •Þ)%œ

where (s :1 • s : 2 )G œ"Þ'%& Ê

(s :1 • s : 2 )G •Þ"

Ê

: 1 ;1 : 2 ;2 € 81 82

: 1 ;1 : 2 ;2 € Þ Substituting for (s : 1 • :s2 )G , 81 82 "Þ'%& Ê

: 1 ;1 : 2 ;2 € • ." 81 82 • .) % œ : 1 ;1 : 2 ;2 € Ê 81 82 ." #Þ%)& œ : 1 ;1 : 2 ;2 € Ê 81 82 The following two assumptions will allow us to calculate the appropriate sample size: " 8 1 œ 8 2 œ 8. 2 The maximum value of : (" • : ) will occur when : œ " • : œ .& . Since values of : 1 and : 2 are unknown, the use of : œ Þ & will provide a valid sample size, although it may be slightly larger than necessary. Then, solving for 8, we obtain #Þ%)& œ

."

Ê.& (.& )(

" " € ) 8 8

Ê

#Þ%)& œ

." .& È#/ 8

È8 œ " ( Þ & ( or 8 œ $ ! ) Þ ( ' . Hence, a common sample size for the researcher's test will be 8 œ $ ! * .

9.59

The hypothesis to be tested is H0 : .1 • .2 œ ! versus Ha : .1 • .2 ž ! and the test statistic is D¸

(B" • B# ) • ! Ë

=21 81

€

=22 82

œ

"!•)

%Þ$ &Þ( Ê € %! %!

œ%

The rejection region, with ! œ .!& , is one-tailed or D ž " Þ ' % & and the null hypothesis is rejected. There is sufficient evidence to indicate a difference in the two means. Hence, we conclude that diet I has a greater mean weight loss than diet II.

9.60

No. The agronomist would have to show experimentally that the increase was $ or more bushels per quadrat in order to acheive practical importance.

205

9.61

The hypothesis to be tested is H0 : .1 • .2 œ ! versus Ha : .1 • .2 Á ! and the test statistic is D¸

(B" • B# ) • ! Ë

=21 81

€

=22

œ

82

"*#&•"*!& %! # $! # € Ë "!! "!!

œ%

The rejection region, with ! œ .!& , is two-tailed or lDl ž " Þ * ' and the null hypothesis is rejected. There is a difference in mean breaking strengths for the two cables.

9.62

The hypothesis to be tested is H0 : .1 • .2 œ ! versus Ha : .1 • .2 Á ! and the test statistic is D¸

(B" • B# ) • ! Ë

=21 81

€

=22

œ

82

"")•"!*

"!# )( Ê € '% '%

œ&Þ#$(

Since no value of ! is specified in advance, we calculate p -value œ P(lDl ž & Þ # $ ( Ñ • # Ð !. ! ! # Ñ œ Þ ! ! ! % Þ Since this p-value isless than Þ!" , you can reject H! at the " % level (highly significant)Þ There is a difference in mean stopping time for the two models.

9.63

a

The hypothesis to be tested is H0 : .1 • .2 œ ! versus Ha : .1 • .2 ž !

and the test statistic is D¸

(B" • B# ) • ! Ë

=21 81

€

=22

œ

82

#%!•##(

*)! )#! Ê € #!! #!!

œ%Þ$$

The rejection region, with ! œ .!& , is one-tailed or D ž " Þ ' % & and the null hypothesis is rejected. There is a difference in mean yield for the two types of spray. b An approximate *& % confidence interval for .1 • .2 is approximately =1 =2 _ _ (B1 • B2 ) „"Þ*' Ë € 81 82 *)! )#! (# % ! • # # ( ) „"Þ*' Ê € #!! #!! " $ „ & Þ ) ) or (Þ"# • (.1 • .2 ) • ")Þ)) 2

9.64

2

a Let : 1 be the proportion of cells in which RNA developed normally when treated with a .' micrograms per millimeter concentration of Actinomysin-D, and : 2 be the proportion of normal cells treated with the higher concentration of Actinomysin-D. The hypothesis to be tested is H0 : : 1 • : 2 œ ! Calculate s :1 œ

versus

Ha : : 1 • : 2 Á !

&& #$ B1 € B2 &&€#$ œÞ()'ß :s2 œ œ Þ $ # * , and s :œ œ œ .&&(Þ The test statistic is (! (! 81 € 82 (!€(!

then

206

Dœ

s : 1 • :s2

sŠ Ë:;

" " € ‹ 81 82

œ

.( ) ' • .$#* œ&Þ%% È.&&( (.%%$ )(" /( ! € " /( ! )

and the p -value is P(lDl ž & Þ % % Ñ • # (.!!!# ) œ .!!!%Þ b Since : -value •Þ!!!% , it must be less than ! œ .!& , and H0 is rejected. We can conclude that there is a difference in the rate of normal RNA synthesis for cells exposed to the two different concentrations of Actinomysin-D.

9.65

a

The hypothesis to be tested is H0 : . œ & ! ' versus Ha : . Á & ! '

The test statistic is D¸

B•.

=/È8

œ

%**•&!' œ • $Þ$$ #" /È"!!

and the p -value is p-value œ PÐD ž $ Þ $ $ Ñ € PÐD • • $ Þ $ $ Ñ œ # (.!!!% ) œÞ!!!) Since the p -value, .!!!) , is smaller than ! œÞ!&ß H0 canbe rejected and we can conclude that the average verbal score for California students in #!!" is different from the national average. b The hypothesis to be tested is H0 : . œ & " % versus Ha : . Á & " % The test statistic is D¸

B•.

=/È8

œ

&"'•&"% œ" #! /È"!!

and the p -value is p-value œ PÐD ž " Ñ € PÐD • • " Ñ œ # (."&)( ) œÞ$"(% Since the p -value, .$"(% , is greater than ! œÞ!&ß H0 cannotbe rejected and we cannot conclude that the average math score for California students in #!!" is different from the national average. c Since the same students are used to measure verbal and math scores, there would not be two independent samples, and the two sample D-test would not be appropriate.

9.66

a The hypothesis to be tested is H0 : . œ ' ' & versus Ha : . Á''&Þ b-c The test statistic is D¸

B•.

=/È8

œ

'&#•''& œ • #Þ&'* $# /È%!

and the Large-Sample Test of a Population Mean applet gives : -value œÞ!"!#Þ c The null hypothesis can be rejected at the & % level but not at the " % level of significance.

9.67

a The parameter of interest is ., the average daily wage of workers in a given industry. A sample of 8 œ % ! workers has been drawn from a particular company within this industry and B, the sample average, has been calculated. The objective is to determine whether this company pays wages different from the total industry. That is, assume that this sample of forty workers has been drawn from a hypothetical population of workers. Does this population have as an average wage . œ & % , or is . different from &% ? Thus, the hypothesis to be tested is

207

H0 : . œ & % versus Ha : . Á & % b-c The test statistic is D¸

B•.

=/È8

œ

&"Þ&!•&% œ • "Þ$$" ""Þ)) /È%!

and the Large-Sample Test of a Population Mean applet gives : -value œÞ")$#Þ (Using Table $ will produce a : value of Þ")$'ÞÑ d Since ! œ .!" is smaller than the p -value, .")$# , H0 cannot be rejected and we cannot conclude that the company is paying wages different from the industry average. e Since 8 is greater than $! , the Central Limit Theorem will guarantee the normality of B regardless of whether the original population was normal or not.

9.68

9.69

The Power of a z-Test applet displays the power correct to only two-decimal places, while the true mean . is only displayed to the nearest integer. Therefore, as you move the slider at the bottom of the applet, several different power probabilities will appear while . remains the same. The power values shown in the applet are all consistent with the actual values given in Table *Þ# . . )(! )(& ))! ))& )*! Table *Þ# Þ*#!( Þ$)*( Þ!&!! Þ$)*( Þ*#!( Applet Þ*! to .*% Þ$$ to .%& Þ!& Þ$$ to .%& Þ*! to .*% _ a In Example *Þ) , the critical values of B that specify the acceptance region were found to be )(%Þ") and ))&Þ)# and " œ P()(%Þ") Ÿ B Ÿ ) ) & Þ ) # Ñ for various alternative values of . . In this exercise, the sample size has been reduced to 8 œ $ ! . Although the procedure remains the same, the value for B G is changed _ because the standard deviation of the random variable is 5/È8 œ # " /È$ ! œ $ Þ ) $ % . The critical values of B corresponding to the acceptance region • " Þ * ' Ÿ D Ÿ " Þ * ' are calculated as _ _ B1 •))! B2 •))! œ • "Þ*' and œ"Þ*' $Þ)$% $Þ)$% _ _ B1 œ)(#Þ%)& and B2 œ))(Þ&"& Remember that the corresponding D-value must be calculated in each case, so that normal curve areas may be _ obtained. When . œ ) ( ! , " œ P Ð)(#Þ%)& Ÿ B Ÿ ) ) ( Þ & " & Ñ and the D -values corresponding to B œ)(#Þ%)& 1 _ and B2 œ))(Þ&"& are D1 œ Then

)(#Þ%)&•)(! œÞ'& $Þ)

%$D2 œ

))(Þ&"&•)(! œ%Þ&( $Þ)

%$" œ PÐ.' & Ÿ D Ÿ % Þ & ( ) œ " • (. % # # œ #. & ( ) and the power is " • " œÞ(%##

b When_ 8 œ ( ! ß the standard deviation of the random variable is 5/È8 œ # " /È( ! œ # Þ & " ! . The critical values of B corresponding to the acceptance region • " Þ * ' Ÿ D Ÿ " Þ * ' are calculated as _ _ B1 •))! B2 •))! œ • "Þ*' and œ"Þ*' #Þ&" #Þ&" _ _ B1 œ)(&Þ!) and B2 œ))%Þ*# When . œ ) ( !, " œ PÐ)(&Þ!) Ÿ B Ÿ ) ) % Þ * # Ñ œ PÐ#Þ!# Ÿ D Ÿ & Þ * % ) œ " • Þ * ( ) $ œ .!#"( and the power is " • " œÞ*()$ c The applet shows power values between Þ(! and .() for . œ ) ( ! Ðcorrect to the nearest integer) when 8 œ $ ! and a power of Þ*) when 8 œ ( ! Þ These are consistent with the hand calculations. d When the sample size is increased, so is the power of the test. 208

9.70

Move the appropriate sliders on the Power of a z-Test applet. As the Sample size and the distance between the null and alternative values of . increase, so does the power. If you decrease !, you will increase ", and hence the power will decrease.

Case Study: An Aspirin a Day...? 1

Let : 1 be the proportion of American physicians who have heart attacks among the hypothetical population of all people who could be treated with &!! mg of aspirin daily, and let : 2 be the proportion of American physicians who have heart attacks among the hypothetical population of all people who could be treated with a placebo. The hypothesis to be tested is H0 : : 1 • : 2 œ !

versus

Ha : : 1 • : 2 Á !

"!% ")* Calculate s :1 œ œÞ!!*%##)&ß :s2 œ œÞ!"("#))(% , and ""!$( ""!$% B1 € B2 "!%€")* s :œ œ œ .!"$#(&$$*Þ The test statistic is then 81 € 82 ""!$(€""!$% Dœ

s : 1 • :s2

:;Š Ëss

" " € ‹ 81 82

œ

.!!*%##)&• .!"("#))(% œ • %Þ%" È.!"("#))(% (.*)#)(""#' )(" /""!$(€"""!$% / )

which is highly significant, with p -value less than .!!!% . 2

Let : 1 be the proportion of British physicians who have heart attacks among the hypothetical population of all people who could be treated with &!! mg of aspirin daily, and let : 2 be the proportion of British physicians who have heart attacks among the hypothetical population of all people who could be treated with a placebo. The hypothesis to be tested is H0 : : 1 • : 2 œ !

versus

Ha : : 1 • : 2 Á !

"'* )) Calculate s :1 œ œÞ!%*#)&&!'ß :s2 œ œÞ!&"%'"*)) , and $%#* "("! B1 € B2 "'*€)) s :œ œ œ .!&!!!*($Þ The test statistic is then 81 € 82 $%#*€"("! s : 1 • :s2

.!%*#)&&!'• .!&"%'"*))

È.!&!!!*($ (.*%***!#( )(" /$ % # * € " "/ ( " ! Ñ " " :;Š € ‹ Ëss 81 82 which is not significant, with p -value, Dœ

œ

œ • Þ

%$PÐD ž .$ % Ñ € PÐD • • Þ $ % Ñ œ # Þ( $ ' ' * ) œ .($$) 3

The results of parts 1 and 2 are very different. There are several reasons for this: a The model was a self-selecting one, not a random sample. Therefore, the samples may not represent the whole population of interest. b The U.S. study has a very low mortality rate compared to the British study. c Other risk factors, such as family history, smoking, hypertension, cholesterol levels, and so on are not reported. These factors could be biasing the results. d The American study was conducted as a double blind study, while the British study was not. Hence, the results may again be biased. The results of the two tests imply that perhaps other factors are at work in this situation. A decision should not be made until the other factors are more thoroughly studied.

209