Week 7: The t Distribution, Confidence Intervals and Tests

Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean

To this point, when examining the mean of a population we have always assumed that the population standard deviation (σ) was known. In practice this is seldom the case. We usually must estimate the population standard deviation with the sample standard deviation s. When we do this, the sampling distribution of the sample mean is no longer normally distributed, because of the adjustment for estimating σ with s. Thus, instead of using the Z , the standard normal distribution, we must use a different distribution called the t distribution.

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Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean The t distributions

Although there is only one Z distribution, there are many, many t distributions. In fact, there is a different t distribution for each sample size used. The shape of each t distribution is very similar to the Z distribution. The shape is bell-shaped, the mean (center) is 0, but the spread is larger. Since we are now estimating the true variance (standard deviaton), we must compensate by using a distribution with a larger spread (more variability to compensate for guessing). The larger the sample size, however, the better our estimate and the closer the t distribution is to the Z distribution.

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Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean The t distributions

The way we distinguish between various t distributions is by finding the degrees of freedom (df ) that correspond to the sample size. For the one sample case, the degrees of freedom are the sample size minus one: df = n − 1. We say that the one-sample t-statistic tn−1 =

x¯ − µ √ s/ n

has the t distribution with n − 1 degrees of freedom. Notice the only difference between the t and the z is that we use s instead of σ.

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Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean The t distributions

t distribution critical values can be found at http://www.stat.tamu.edu/stat30x/zttables.php#ttable. The degrees of freedom are listed in the left column. The individual t-values are inside the table rather than the row/column labels as in the Z table. The probabilities on top (column headings) are areas to the right of the t-values within the table whereas the probabilities within the Z table are areas to the left. Confidence levels are given at the bottom of the table. Make sure to get acquainted with this table and how it differs from the Z table. 5 / 51

Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean The t distributions

Differences between Z and t tables:

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Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean The t distributions

How the Z and t distributions compare:

From http://www.stat.tamu.edu/˜jhardin/applets/signed/T.html As sample size, n, increases, the t distribution converges to Z distribution. 7 / 51

Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean The t distributions

Changing from σ to s forces us to change from z to tn−1 (for the one sample case), but the steps in producing confidence intervals and hypothesis tests are the same as we have seen previously. Remember from Week 1, s is calculated from the data using the formula:

v u u s=t

n 1 X (xi − x¯ )2 n−1 i=1

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Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean The one-sample Confidence Interval (CI) for µ with Unknown σ

The formula for a confidence interval for µ with unknown σ is s x¯ ± tα/2,n−1 √ n tα/2,n−1 is found in the t table. It must correspond to the appropriate df = n − 1 row and the correct α-level column label at the bottom. It is easier to find the confidence level at the bottom of the table and go up to the correct df .

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Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean The one-sample Confidence Interval (CI) for µ with Unknown σ

One-sample t Confidence Interval Example An economist wants to determine the annual average amount that a family of four in the United States spends on housing. He randomly selects 85 families of size four and finds the amount they spent on housing the previous year. The economist wishes to estimate the mean with 99% confidence. Information we have: Sample Size: n = 85. Data: $6,789, $8,233, $4,784, · · · , $5,974 (85 numbers) Calculated from the data: x¯ = $6, 219, s = $1,978 Degrees of Freedom: df = n − 1 = 85 − 1 = 84 10 / 51

Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean The one-sample Confidence Interval (CI) for µ with Unknown σ

One-sample t CI Example 1, 978 s x¯ ± tα/2,n−1 √ = 6, 219 ± 2.639 √ n 85 = (5652.82, 6785.18)

tα/2,n−1 is found in the t table. We first go to the 99% confidence level at the bottom. Since 84 is NOT in the table, we then go up to 80 df (always round down = up in the table). Thus, t0.005,80 = 2.639. This is a 99% confidence interval for the true average amount a family of four in the United States spends on housing annually.

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Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean Test of Hypotheses Procedure for a 1-sample t test:

Testing Procedure:

1

State the null hypothesis, H0 .

2

State the alternative hypothesis, HA .

3

Determine the level of significance (e.g., α = 0.05).

4

Calculate the test statistic (t-statistic): tn−1 =

x¯ − µ0 √ s/ n

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Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean Test of Hypotheses Procedure for a 1-sample t test:

5

Determine the range of the p-value from the t table. For a greater than test: HA : µ > µ0 p-value = P(T > t) For a less than test: HA : µ < µ0 p-value = P(T < t) For a two-sided test: HA : µ 6= µ0 p-value = P(T ≥ |t| or T ≤ −|t|) = 2P(T ≥ |t|) We cannot determine the exact p-value unless we use a computer, so it is more common to state a range for the p-value.

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Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean Test of Hypotheses Procedure for a 1-sample t test:

6

Reject or fail to reject H0 based on the p-value. If the p-value is less than or equal to α, reject H0 . It the p-value is greater than α, fail to reject H0 .

7

State your conclusion. If H0 is rejected with the two-sided alternative, “There is significant statistical evidence that the population mean is different from µ0 .” If H0 is not rejected, “There is NOT significant statistical evidence that the population mean is different than µ0 .” Notice that the steps are exactly the same as for the case where σ is known except for the test statistic formula and the p-value.

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Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean Test of Hypotheses Procedure for a 1-sample t test:

TV Example: Suppose that the data collected from a class survey is a random sample from the entire university (which obviously is not). We wish to see if there is evidence that the average amount of television watched for students here is more than 7 hours per week.

Sample Size: n = 38 x¯ : 8.05, s = 7.46 Degree of Freedom: df = 37

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Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean Test of Hypotheses Procedure for a 1-sample t test:

TV Example con’t: State the null hypothesis: H0 : µ = 7 State the alternative hypothesis: HA : µ>7 State the level of significance: α = 0.05 Calculate the test statistic. tn−1 =

x¯ − µ0 √s n

=

8.05 − 7 7.46 √ 38

= 0.868

Find the p-value. p-value = P(T ≥ t) = P(T ≥ 0.868)

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Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean Test of Hypotheses Procedure for a 1-sample t test:

TV Example con’t: Since there isn’t a 37th row, we use 30. Our test statistic is 0.868 which falls between 0.854 and 1.055. So according to the top of the table, our p-value falls between 0.20 and 0.15. Note: since this is a greater than test and the areas at the top are areas greater than the corresponding t-values, we did not subtract from 1 nor multiply by 2. Do we reject or fail to reject H0 based on the p-value? α = 0.05 < 0.15 < p-value < 0.20, so we fail to reject H0 . State the conclusion: “There is NOT significant statistical evidence that the average amount of television watched is more than 7 hours per week at the 0.05 level of significance.”

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Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean - Dependent Two Samples (Matched Pairs)

To this point, we have only looked at tests for a single sample. Soon we will look at confidence intervals and hypothesis tests for comparing two groups. There is a special 2-sample case in which each individual can be given both treatments. These two samples are not independent. We can reduce the two samples to a single sample using a matched pairs design and eliminate one source of variability. To analyze matched pairs data, we first reduce the data from two samples to one sample by taking the difference for each individual and then use the 1-sample test.

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Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean - Dependent Two Samples (Matched Pairs) Example of matched pairs design

Matched Pairs Examples: Students are each given a pre-test and a post-test to determine the amount of material learned in a given time interval. To examine the effect of a new drug, a large group of identical twins is identified. One twin is given a treatment and the other a placebo. An ophthalmologist is examining the importance of the dominant eye in reading. A large group of subjects is asked to read a passage with dominant eye covered and again with the non-dominant eye covered. It can be seen in each of these examples that something pairs the two responses.

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Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean - Dependent Two Samples (Matched Pairs) Example of matched pairs design

The data is reduced from two samples to one by subtracting one of the responses from the other. We could subtract each pre-test score from each post-test score. We could subtract each placebo response from each treatment response. We could subtract the time taken to read the passage with the non-dominant eye from that with the dominant eye.

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Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean - Dependent Two Samples (Matched Pairs) Matched Pairs Confidence Interval

After reducing the data to a single sample, we use the same formula as for a confidence interval for µ with unknown σ, namely, sd x¯d ± tα/2,nd −1 √ nd where x¯d and sd are calculated from the differences, nd is the number of differences, and tα/2,nd −1 is from t-table with df = nd − 1. Note that this nd is actually half the total number of observations since there are two observations per individual.

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Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean - Dependent Two Samples (Matched Pairs) Matched Pairs Confidence Interval

Golf Ball Example: “In the manufacture of golf balls two procedures are used. Method I utilizes a liquid center and method II, a solid center. To compare the distance obtained using both types of balls, 12 golfers are allowed to drive a ball of each type, and the length of the drive (in yards) is measured.” (from Milton, McTeer, and Corbet, Introduction to Statistics, 1997) The manufacturer wants to estimate the mean difference with 90% confidence.

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Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean - Dependent Two Samples (Matched Pairs) Matched Pairs Confidence Interval

Golf Ball Example con’t: Sample Size: nd = 12 x¯d = 9.52 sd = 3.12 df = nd − 1 = 11

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Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean - Dependent Two Samples (Matched Pairs) Matched Pairs Confidence Interval

Golf Ball Example con’t: 3.12 sd x¯d ± tα/2,nd −1 √ = 9.52 ± 1.796 √ = (7.90, 11.14) nd 12

To find t0.05,11 , we first go to the 90% confidence level at the bottom. Then we go up to find df = 11. Thus, t0.05,11 = 1.796. This is a 90% confidence interval for the true average difference for the distance traveled for the two types of golf balls.

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Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean - Dependent Two Samples (Matched Pairs) Paired t test

Hypotheis Test for Paired Data: Again reduce the data to a single sample, then use the 1-sample t-test. tnd −1 =

x¯d − µd s √d nd

Again, note that the degrees of freedom are from the reduced one sample.

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Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean - Dependent Two Samples (Matched Pairs) Paired t test

Keyboard Example: “Suppose we want to compare two brands of computer keyboards, which we will denote as keyboard 1 and keyboard 2. Keyboard 1 is a standard keyboard, while keyboard 2 is specially designed so that the keys need very little pressure to make them respond. The manufacturer of keyboard 2 would like to claim that typing can be done faster using keyboard 2. A simple random sample of n = 30 teachers was selected from a population of high-school teachers attending a national conference. Each teacher typed the same page of text once using keyboard 1 and once using keyboard 2. For each teacher the order in which the keyboards were used was determined by the toss of a coin. The variable measured was the time (in seconds) for each teacher to correctly type the page of text · · · ” (from Graybill, Iyer and Burdick, Applied Statistics, 1998). 26 / 51

Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean - Dependent Two Samples (Matched Pairs) Paired t test

Keyboard Example con’t:

Sample Size: nd = 30 x¯d = x¯2 − x¯1 = −3.53 sd = 8.56 df = nd − 1 = 29

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Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean - Dependent Two Samples (Matched Pairs) Paired t test

Keyboard Example con’t: State the null hypothesis: H0 : µd = µ2 − µ1 = 0 State the alternative hypothesis: HA : µd < 0 Why less than? Determine the level of significance: α = 0.05 Calculate the test statistic: t29 =

x¯d − µd s √d nd

=

−3.53 − 0 8.56 √ 30

= −2.26

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Week 7: The t Distribution, Confidence Intervals and Tests Inference for a Single Mean - Dependent Two Samples (Matched Pairs) Paired t test

Keyboard Example conclusion: Find the p-value. p − value = P(T ≤ t) = P(T ≤ −2.26) = between 0.01 and 0.02 Use degree of freedom as 29. 0.01 < p-value < 0.02 < α = 0.05. Therefore, we reject H0 and conclude HA is true. State the conclusion: “There is significant statistical evidence that the average amount of time needed to type the passage is lower for keyboard 2 than keyboard 1 at the 0.05 level of significance.”

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Week 7: The t Distribution, Confidence Intervals and Tests Comparing Two Means-Independent Two Samples w/ σ1 6= σ2

Two Independent Populations Means What about comparing two samples or populations? First, we can compare the samples graphically. Histograms or stemplots for each. The easiest way is side-by-side boxplots since the datasets are put on the same scale. We can also compare means, medians, standard deviations, etc. But we can’t make any conclusions without statistical evidence.

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Week 7: The t Distribution, Confidence Intervals and Tests Comparing Two Means-Independent Two Samples w/ σ1 6= σ2

Two Independent Populations Means To compare two independent population means, we test the difference between the means. But we need the sampling distribution for the difference. Suppose X ∼ N(µX , σX2 ) and Y ∼ N(µY , σY2 ) are independent. Then µX ±Y = µX ± µY and σX ±Y = X ±Y

∼

N

q σX2 + σY2 . In particular,

q 2 ! 2 2 µX ± µY , σX + σY

Remember, you must find the variance first and then take the square root to get the standard deviation. So whether we want the sum or the difference, we always SUM the variances and then take the square root. 31 / 51

Week 7: The t Distribution, Confidence Intervals and Tests Comparing Two Means-Independent Two Samples w/ σ1 6= σ2

For example: Let X and Y denote the average scores on the first test for two different sections, respectively. (In reality, these probably aren’t independent if they cover the same material.) It is known that X

∼

N(80, 42 )

Y

∼

N(70, 32 )

Since the same formulas work for means:  p 2  √ 2 X − Y ∼ N 80 − 70, 42 + 32 = N(10, 5 ) If not given the distribution of the means, only the individuals, remember the variance for the mean, σX2¯ = σX2 /n. 32 / 51

Week 7: The t Distribution, Confidence Intervals and Tests Comparing Two Means-Independent Two Samples w/ σ1 6= σ2

Two Independent Populations Means When we are interested in comparing two population means and we are estimating the population standard deviations σ1 and σ2 with s1 and s2 , the conservative two-sample t-statistic is then t=

(x¯1 − x¯2 ) − (µ1 − µ2 ) q s12 /n1 + s22 /n2

with degrees of freedom equal to the smaller of n1 − 1 and n2 − 1, i.e., df = minn1 − 1, n2 − 1.

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Week 7: The t Distribution, Confidence Intervals and Tests Comparing Two Means-Independent Two Samples w/ σ1 6= σ2 Two Sample t-tests

The null hypothesis can be any of the following: H 0 : µ1 = µ2

H 0 : µ1 ≤ µ2

H0 : µ1 ≥ µ2

The alternative hypothesis can be any of the following: HA : µ1 6= µ2

H A : µ1 > µ 2

H A : µ1 < µ 2

The other steps are the same as those used for the tests we have looked at previously.

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Week 7: The t Distribution, Confidence Intervals and Tests Comparing Two Means-Independent Two Samples w/ σ1 6= σ2 Two Sample t-tests

Tomato Example “There has been some discussion among amateur gardeners about the virtues of black plastic versus newspapers as weed inhibitors for growing tomatoes. To compare the two, several rows of tomatoes are planted. Black plastic is used around nine randomly selected plants and newspaper around the remaining ten. All plants start at virtually the same height and receive the same care. The response of interest is the height in feet after a month-growth.” (from Milton, McTeer, and Corbet, Introduction to Statistics, 1997). Perform a test to see if there is any difference between the average heights with significance level 0.10.

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Week 7: The t Distribution, Confidence Intervals and Tests Comparing Two Means-Independent Two Samples w/ σ1 6= σ2 Two Sample t-tests

Tomato Example

Sample Sizes: n1 = 9, n2 = 10 x¯1 = 1.87, x¯2 = 1.49 s1 = 0.63, s2 = 0.43 df = n1 − 1 = 9 − 1 = 8 because n1 is smaller than n2 .

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Week 7: The t Distribution, Confidence Intervals and Tests Comparing Two Means-Independent Two Samples w/ σ1 6= σ2 Two Sample t-tests

Tomato Example State the null hypothesis: H0 : µ1 = µ2 ⇒ µ1 − µ2 = 0 State the alternative hypothesis: Ha : µ1 6=µ2 State the level of significance: α = 0.10 Calculate the test statistic. t=

(x¯1 − x¯2 ) − (µ1 − µ2 ) (1.87 − 1.49) − 0 q = p = 1.519 0.632 /9 + 0.432 /10 s12 /n1 + s22 /n2

Note: We will use the computer for this calculation. The formula is only provided for completeness.

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Week 7: The t Distribution, Confidence Intervals and Tests Comparing Two Means-Independent Two Samples w/ σ1 6= σ2 Two Sample t-tests

Tomato Example Find the p-value. p − value = 2P(T ≥ |t|) = 2P(T ≥ 1.519) = 2(between 0.05 and 0.10) = between 0.10 and 0.20 Use df = 8. α = 0.10 < p-value < 0.20, so we fail to reject H0 State the conclusion: “There is not significant statistical evidence that the average tomato plant heights are different for the two types of weed inhibitors at the 0.10 level of significance.”

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Week 7: The t Distribution, Confidence Intervals and Tests Comparing Two Means-Independent Two Samples w/ σ1 6= σ2 Two-Sample t Confidence Interval

Confidence Intervals The confidence interval for the difference of two population means (µ1 − µ2 ) is s (x¯1 − x¯2 ) ± tα/2,df

s12 s2 + 2 n1 n2

Where tα/2,df corresponds to the desired confidence level and df = min{n1 -1, n2 -1}.

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Week 7: The t Distribution, Confidence Intervals and Tests Comparing Two Means-Independent Two Samples w/ σ1 6= σ2 Two-Sample t Confidence Interval

Commercial Example: “There is some concern that TV commercial breaks are becoming longer. The observations on the following slide are obtained on the length in minutes of commercial breaks for the 1984 viewing season and the current season.” (from Milton, McTeer, and Corbet, Introduction to Statistics, 1997) Find a 95% confidence interval for the difference between the true averages of the two seasons.

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Week 7: The t Distribution, Confidence Intervals and Tests Comparing Two Means-Independent Two Samples w/ σ1 6= σ2 Two-Sample t Confidence Interval

Commercial Example con’t: Sample Sizes: n1 = 16, n2 = 16 x¯1 = 2.01, x¯2 = 2.36 s1 = 0.49, s2 = 0.19 df = n1 − 1 = 16 − 1 = 15 because n1 and n2 are same. t0.025,15 = 2.131. Go to the 95% confidence level at the bottom. Then go up to df = 15. 41 / 51

Week 7: The t Distribution, Confidence Intervals and Tests Comparing Two Means-Independent Two Samples w/ σ1 6= σ2 Two-Sample t Confidence Interval

Commercial Example con’t: s r s22 s12 0.492 0.192 + = (2.01 − 2.36) ± 2.131 (x¯1 − x¯2 ) ± t0.025,15 + n1 n2 16 16 = (−0.63, −0.07)

This is a 95% confidence interval for the true difference of average length in minutes for commercials between 1984 and the present. At the 5% significance level, we could conclude that there is a difference in the average length for commericals between 1984 and the present. Why?

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Week 7: The t Distribution, Confidence Intervals and Tests Comparing Two Means - Independent Two Samples w/ σ1 = σ2 Pooled Estimator

Added Assumption of Equal Variances Previously, we discussed two-sample t procedures from two populations with two unknown standard deviations. We then used the sample standard deviations to estimate the population standard deviations. But what if the two populations have the same standard deviation? This estimate is called the pooled estimator of sp2 because it combines (pools) the information in both samples. sp2 =

(n1 − 1)s12 + (n2 − 1)s22 n1 + n2 − 2

The pooled standard deviation is the average of the two based on each sample size. This gives us a better estimate for the true population standard deviation since it is based on a larger sample = n1 + n2 (more data). 43 / 51

Week 7: The t Distribution, Confidence Intervals and Tests Comparing Two Means - Independent Two Samples w/ σ1 = σ2 Test Statistics

Suppose that an SRS of size n1 is drawn from a normal population with unknown mean µ1 and that an independent SRS of size n2 is drawn from another normal population with unknown mean µ2 . Suppose further that we know that the two populations have the SAME standard deviation. Thus, the pooled two-sample t statistic is tn1 +n2 −2 =

sp

p

x¯1 − x¯2 1/n1 + 1/n2

The degrees of freedom are df = n1 + n2 - 2 due to the larger pooled sample. The added assumption for the pooled t-test makes it more powerful than the 2-sample t-test which means it is easier to detect a false H0 . 44 / 51

Week 7: The t Distribution, Confidence Intervals and Tests Comparing Two Means - Independent Two Samples w/ σ1 = σ2 HT and CI

Hypothesis test procedures are the same. Only the test statistic with df = n1 + n2 − 2 and p-value are different. A level 1 − α confidence interval for µ1 − µ2 is r (x¯1 − x¯2 ) ± tα/2,n1 +n2 −1 sp

1 1 + n1 n2

Where tα/2,n1 +n2 −1 corresponds to the desired confidence level and df = n1 + n2 - 2. Again, all of these calculations will be done with the computer, NEVER by hand.

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Week 7: The t Distribution, Confidence Intervals and Tests Comparing Two Means - Independent Two Samples w/ σ1 = σ2 HT and CI

Tomato Example Revisited: Still using the tomato example aforementioned, but now assuming the two populations have the same standard deviation, i.e. assume that σ1 = σ2 . Sample Sizes: n1 = 9, n2 = 10 x¯1 = 1.87, x¯2 = 1.49 s1 = 0.63, s2 = 0.43 Note: In practice, we should NOT assume equal variances unless told so or the values of sample variance are very close to each other. The rule of thumb says not more than twice the size. df = n1 + n2 − 2 = 9 + 10 − 2 = 17 (note the df is larger than before).

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Week 7: The t Distribution, Confidence Intervals and Tests Comparing Two Means - Independent Two Samples w/ σ1 = σ2 HT and CI

Tomato Example Revisited: State the null hypothesis: H0 : µ1 = µ2 State the alternative hypothesis: HA : µ1 6= µ2 State the level of significance: α = 0.10 Calculate the test statistic (from the computer). t=

(x¯1 − x¯2 ) − (µ1 − µ2 ) (1.87 − 1.49) − 0 p p = = 1.550 sp 1/n1 + 1/n2 0.53 1/9 + 1/10

Since the sample standard deviations were close, this value isn’t much different than the 2-sample t = 1.519.

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Week 7: The t Distribution, Confidence Intervals and Tests Comparing Two Means - Independent Two Samples w/ σ1 = σ2 HT and CI

Tomato Example Revisited: The p-value, however comes from a different row of the t table, df = 17 vs. df = 8. p − value = 2P(T ≥ |t|) = 2P(T ≥ 1.550) = 2(between 0.05 and 0.10) = between 0.10 and 0.20 In this case though, the p-value range is the same. From the computer, the true p-values are 0.1395 are 0.1527. The pooled t-test will always have a smaller p-value and therefore it would be easier to reject. α = 0.10 < p-value < 0.20, so we fail to reject H0 . State the conclusion: the same as before. 48 / 51

Week 7: The t Distribution, Confidence Intervals and Tests Comparison of the Two Sample Tests Power vs. Conservatism

With every added assumption, we get more power. The z-test is more powerful than the t-test since we assume the variances are known. The pooled t-test is more powerful than the 2-sample t-test since we assume that the unknown variances are equal. The larger the df , the more powerful the test. Notice that the z is the last row of the t table. It has the largest possible sample, the whole population. To be conservative means to use a smaller df and so we get larger p-values and wider confidence intervals. If we reject with one df , we would also reject with a larger df (more data) assuming that H0 is false. 49 / 51

Week 7: The t Distribution, Confidence Intervals and Tests Comparison of the Two Sample Tests Power vs. Conservatism

The paired t-test is even more powerful than the pooled t-test, but not because of its degrees of freedom. In the paired t-test procedure, we reduce the total variance by eliminating the difference in the individuals. There are (at least) two sources of variability in any two sample situation: the difference due to the different population means and the difference due to the different individuals. The smaller the variance, the easier it is to see a difference in means. Remember s or σ is always in the denominator. The smaller the denominator, the larger the test statistics. The larger the test statistic, the smaller the p-value and the more likely we are to reject. 50 / 51

Week 7: The t Distribution, Confidence Intervals and Tests Comparison of the Two Sample Tests Power vs. Conservatism

Example: We want to know if retaking the SAT will improve your score. There is the difference in the test scores for each person (paired difference). It is unlikely you will make exactly the same score, but ‘did everyone improve’ is the question? There is also a difference in people’s scores. Some people will do well both times and some will not do as well either time. (Did the person sitting next to you get the same score?) The paired t-test gets rid of the difference in individuals by only comparing ‘within’ an individual. But remember, this test requires paired, dependent data. Often we cannot do this due to time restraints, etc. 51 / 51