Questions Before One Sample Proportion Inference Answers
Questions Before One Sample Proportion Inference Answers 1. Suppose you take a SRS of size n from a population and calculate the proportion of the sam...
Questions Before One Sample Proportion Inference Answers 1. Suppose you take a SRS of size n from a population and calculate the proportion of the sample to be pˆ . a. Write a general form for a 96% confidence interval. " pˆ (1 − pˆ ) % pˆ ± 2.054 $ ' n # & b. What conclusion about the population can you draw from the confidence interval? Why? 
We are 96% confident that an interval created as shown above will capture he population proportion

96% of all intervals like the one shown above will capture the population proportion
The reason why these two description are valid is because 96% of all the sample proportions are within 2.054 standard deviations from the population proportion therefore if we go 2.054 standard deviations from a given sample proportion we can use one of the above conclusions. c. What would happen to the margin of error and therefore the confidence interval if we were to change size of the sample? Why? We know that as the sample size increases the standard deviation of the proportions decreases. Therefore as the sample size increases the margin of error decreases so the width of the confidence interval gets smaller. If the sample size decreases the margin of error increases so the width of the confidence interval gets larger. d. What would happen to the margin of error and therefore the confidence interval if we were to change the Clevel? Why? If we increase the confidence level we increase the size of the critical value and therefore increase the margin of error and increase the width of the confidence interval. If we decrease the confidence level we decrease the size of the critical value and therefore decrease the margin of error and decrease the width of the confidence interval.
2. What is the value of z* when you are creating a 92% confidence interval? z* = 1.751
3. What is the value of z* when you are creating a 85% confidence interval? z* = 1.44
4. What is the value of z* when you are performing a onetail test and α = 0.03? z* = 1.88
5. What is the value of z* when you are performing a onetail test and α = 0.02? z* = 2.054
6. What is the value of z* when you are performing a twotail test and α = 0.07? z* = 1.812
7. What is the value of z* when you are performing a twotail test and α = 0.12? z* = 1.555
8. If α = 0.05 then what confidence level would give same result as a: a. OneTail Test 90%
b. TwoTailed Test 95%
9. If the confidence level is 94%, what α value would give same result as a: a. OneTail Test
α = 0.03 b. TwoTailed Test
α = 0.06 10. Your local newspaper polls a random sample of voters. How big of a sample size is required to have a margin of error within 1% when creating a 90% confidence interval.
! (0.5) ( 0.5) $ ! $ & ⇒ 0.00608 = # (0.25 & ⇒ ( 0.00608)2 = 0.25 ⇒ n= 0.25 2 = 6762.898 0.01 =1.645## & n n " n % (0.00608) " % The required sample size to guarantee a margin of error within 1% is 6763
11. A recent study examined a random sample of 384 children and found that 46 of them showed signs of autism a. Determine the 95% confidence interval and clearly express what you have calculated. (0.08731, 0.15227). I am 95% confident that the percentage of children with autism is between 8.731% and 15.227%
b. In the 1980s it was accepted that autism affected about 5% of the nation’s children. What does this confidence interval suggest? Since all the values in the interval (0.08731, 0.15227) are larger than 0.05 I would conclude that the proportion of children diagnosed with autism has increased.
12. A recent study surveyed a random sample of 881 and found that 42% had smoked. P = Proportion of people that have smoked Ho: P = 0.44 Ha: P < 0.44 I will perform a onesample proportion ztest (0.386,0,454). I am 96% confident that the percentage of people who have smoked is between 38.6% and 45.4%. Since 44% lies within the interval we do not have sufficient evidence to conclude that the percent of adults who have smoked in today’s population is less than 44% so we cant conclude the percent of adults who smoke is less than percent who smokes in 1960s
13. Mars Inc., makers of M&M candies, claims that they produce M&M’s with the certain distribution shown below to the left. A bag of M&M’s was randomly selected from a grocery store shelf and the color counts were recorded as follows below to the right. Our goal is to check the company’s claim for the proportion of yellow M&M’s Brown Orange Red
30% Green 10% Yellow 20% Blue
10% 20% 10%
Brown Orange Red
16 5 11
Green Yellow Blue
P = Proportion of Yellow M&M’s Ho: P = 0.20 Ha: P ≠ 0.20 I will perform a onesample proportion ztest (0.1953,0.4277). I am 95% confident that the percentage of yellow M&M’s is between 19.53% and 42.77%. Since 20% lies within the interval I can not conclude that the percentage of yellow M&M’s does not equal 20%
7 19 3
14. A 95% confidnece interval for the proportion of college age students between the are resgistered to vote random sample is (0.573, 0.827) a. Compute the value of the sample proportion, pˆ .
pˆ = 0.7 b. Compute the value of the Margin of Error.
M.E. = 0.127
c. Compute the value of the Standard Error.
S.E. = 0.0648
d. Compute the Sample Size.
0.0648 =
(0.7)(0.3) n
n = 50
e. A claim is made that 73% of college age students are registered to vote. Does this 73% seem reasonable based on the above confidence interval? Since 0.73 lies within the interval we can not rule it out as a possible proportion of college age students registered to vote Therefore we must condier 73% as a reasonable percentage of college age studenst that are registered to vote.
AP Statistics Quiz B – Chapter 19 – Key The countries of Europe report that 46% of the labor force is female. The United Nations wonders if the percentage of females in the labor force is the same in the United States. Representatives from the United States Department of Labor plan to check a random sample of over 10,000 employment records on file to estimate a percentage of females in the United States labor force.
15.
a.
1. The representatives from the Department of Labor want to estimate a percentage of females in the United States labor force to within ±5%, with 90% confidence. How many employment records should they sample? pˆ qˆ ˆˆ ME = z* pq ME z * n n They should sample at least 269 employment records.
(0.5) (0.5) (0.46)(0.54) 0.05 1.645 = 1.645 0.05 n n
They should sample at least 271 employment records
1.645 (0.46)(0.54) 1.645(0.5) nn = 0.05 0.05 268.87 n n = 270.06 269
b.2. They actually select a random sample of 525 employment records, and find that 229 of the people are females. Create the confidence interval. We have a random sample of less than 10% of the employment records, with 229 successes (females) and 296 failures (males), so a Normal model applies. n = 525, pˆ
0.436 and qˆ
margin of error: ME
z*
0.564 , so SE ( pˆ ) SE ( pˆ )
Confidence interval: pˆ ME
ˆˆ pq n
1.645 0.022
0.436 0.564 525 0.0362
0.022
0.436 0.0362 or (0.3998, 0.4722)
c.3. Interpret the confidence interval in this context. We are 90% confident that between 40.0% and 47.2% of the employment records from the United States labor force are female.
d.4. Explain what 90% confidence means in this context. If many random samples were taken, 90% of the confidence intervals produced would contain the actual percentage of all female employment records in the United States labor force.
e.5. Should the representatives from the Department of Labor conclude that the percentage of females in their labor force is lower than Europe’s rate of 46%? Explain. No. Since 46% lies in the confidence interval, (0.3998, 0.4722), it is possible that the percentage of females in the labor force matches Europe’s rate of 46% female in the labor force.