Inferences on Two Samples

Inferences on Two Samples
Chapter 11 Inferences on Two Samples

Inference about Two Population Proportions
Section 11.1 Inference about Two Population Proportions

A sampling method is independent when the individuals selected for one sample do not dictate which individuals are to be in a second sample. A sampling method is dependent when the individuals selected to be in one sample are used to determine the individuals to be in the second sample. Dependent samples are often referred to as matched-pairs samples. It is possible for an individual to be matched against him- or herself.

Parallel Example 1: Distinguish between Independent and
Parallel Example 1: Distinguish between Independent and Dependent Sampling For each of the following, determine whether the sampling method is independent or dependent. A researcher wants to know whether the price of a one night stay at a Holiday Inn Express is less than the price of a one night stay at a Red Roof Inn. She randomly selects 8 towns where the location of the hotels is close to each other and determines the price of a one night stay. A researcher wants to know whether the “state” quarters (introduced in 1999) have a mean weight that is different from “traditional” quarters. He randomly selects 18 “state” quarters and 16 “traditional” quarters and compares their weights.

Solution The sampling method is dependent since the 8 Holiday Inn Express hotels can be matched with one of the 8 Red Roof Inn hotels by town. The sampling method is independent since the “state” quarters which were sampled had no bearing on which “traditional” quarters were sampled.

Parallel Example 1: Distinguish between Independent and
Parallel Example 1: Distinguish between Independent and Dependent Sampling For each of the following, determine whether the sampling method is independent or dependent. Do women tend to select a spouse who has an IQ higher than their own? To answer this question, researchers randomly selected 20 women and their husbands. They measured the IQ of each husband-wife team to determine if there is significant difference in IQ. Suppose students are either enrolled in traditional lecture course or a lab- based format. There were 1200 students enrolled in the traditional lecture format and 300 enrolled in lab-based format. Once the course ended, the researchers looked at grades of students. The goal of the study was to determine whether the proportion of students who passed the lab-based format exceeded that of the lecture format.

Sampling Distribution of the Difference
between Two Proportions (Independent Sample) Suppose that a simple random sample of size n1 is taken from a population where x1 of the individuals have a specified characteristic, and a simple random sample of size n2 is independently taken from a different population where x2 of the individuals have a specified characteristic. The sampling distribution of , where and , is approximately normal, with mean and standard deviation

Sampling Distribution of the Difference
between Two Proportions (Independent Sample) provided that and and each sample size is no more than 5% of the population size.

Sampling Distribution of the Difference between Two Proportions
The standardized version of is then written as which has an approximate standard normal distribution.

The best point estimate of p is called the pooled estimate of p, denoted , where
Test statistic for Comparing Two Population Proportions

the samples are independently obtained using simple random sampling,
Hypothesis Test Regarding the Difference between Two Population Proportions To test hypotheses regarding two population proportions, p1 and p2, we can use the steps that follow, provided that: the samples are independently obtained using simple random sampling, and and n1 ≤ 0.05N1 and n2 ≤ 0.05N2 (the sample size is no more than 5% of the population size); this requirement ensures the independence necessary for a binomial experiment.

Step 1: Determine the null and alternative hypotheses
Step 1: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways:

Classical Approach Step 2: Compute the test statistic where

Classical Approach Step 3: Compare the critical value with the test statistic: Step 4: State the conclusion.

P-Value Approach Step 3: Find P-value using technology or standard table. If P-value < α, reject the null hypothesis. Step 4: State the conclusion.

Parallel Example 1: Testing Hypotheses Regarding Two
Parallel Example 1: Testing Hypotheses Regarding Two Population Proportions An economist believes that the percentage of urban households with Internet access is greater than the percentage of rural households with Internet access. He obtains a random sample of 800 urban households and finds that 338 of them have Internet access. He obtains a random sample of 750 rural households and finds that 292 of them have Internet access. Test the economist’s claim at the α = 0.05 level of significance.

Solution We must first verify that the requirements are satisfied:
The samples are simple random samples that were obtained independently. x1=338, n1=800, x2=292 and n2=750, so The sample sizes are less than 5% of the population size.

H0: p1 - p2=0 versus H1: p1 - p2 > 0
Solution Step 1: We want to determine whether the percentage of urban households with Internet access is greater than the percentage of rural households with Internet access. So, H0: p1 = p2 versus H1: p1 > p2 or, equivalently, H0: p1 - p2=0 versus H1: p1 - p2 > 0 The level of significance is α = 0.05.

Solution Step 2: The pooled estimate of is: The test statistic is:

Solution: Classical Approach
This is a right-tailed test with α = The critical value is z0.05=1.645.

Step 3: Since the test statistic, z0=1.33 is less than the critical value z.05=1.645, we fail to reject the null hypothesis.

Solution: P-Value Approach
Because this is a right-tailed test, the P-value is the area under the normal to the right of the test statistic z0= That is, P-value = P(Z > 1.33) ≈ 0.09.

Solution: P-Value Approach
Step 3: Since the P-value is greater than the level of significance α = 0.05, we fail to reject the null hypothesis.

Solution Step 4: There is insufficient evidence at the α = level to conclude that the percentage of urban households with Internet access is greater than the percentage of rural households with Internet access.

Example In trials, 3774 adult and adolescent allergy patients were randomly divided into two groups. The patients in group 1 (experimental group) received 200 𝜇g of Nasonex, while patients in group 2 (control group) received a placebo. Of the 2013 patients in group one, 547 reported headaches as a side effect. Of the 1671 patients in the control group, 368 reported headaches as a side effect. Is there significant evidence to conclude that the proportion of Nasonex users who experienced headaches as a side effect is greater than the proportion in the control group at the 𝛼=0.05 level of significance.

Constructing a (1 – α)•100% Confidence Interval for the Difference between Two Population Proportions To construct a (1 – α)•100% confidence interval for the difference between two population proportions, the following requirements must be satisfied: the samples are obtained independently using simple random sampling, , and n1 ≤ 0.05N1 and n2 ≤ 0.05N2 (the sample size is no more than 5% of the population size); this requirement ensures the independence necessary for a binomial experiment.

Constructing a (1 – α)•100% Confidence Interval for the Difference between Two Population Proportions Provided that these requirements are met, a (1 – α)•100% confidence interval for p1 – p2 is given by Lower bound: Upper bound:

Parallel Example 3: Constructing a Confidence Interval for
Parallel Example 3: Constructing a Confidence Interval for the Difference between Two Population Proportions An economist obtains a random sample of 800 urban households and finds that 338 of them have Internet access. He obtains a random sample of 750 rural households and finds that 292 of them have Internet access. Find a 99% confidence interval for the difference between the proportion of urban households that have Internet access and the proportion of rural households that have Internet access.

Solution We have already verified the requirements for constructing a confidence interval for the difference between two population proportions in the previous example. Recall

Solution Thus, Lower bound = Upper bound =

Solution We are 99% confident that the difference between the proportion of urban households that have Internet access and the proportion of rural households that have Internet access is between –0.03 and Since the confidence interval contains 0, we are unable to conclude that the proportion of urban households with Internet access is greater than the proportion of rural households with Internet access.

Example Gallup surveyed 1100 adult Americans on May 6-9, 2002, and conducted an independent survey of 1100 adult Americans on May 8-11, In both surveys they asked the following: “Right now, do you think the state of moral values in the country as a whole is getting better or getting worse?” On May 8-11, 2014, 816 of the 1100 surveyed responded that the states of moral values is getting worse; on May 6-9, 2002, 737 of the 1100 surveyed responded that the state of moral values is getting worse. Construct and interpret a 90% confidence interval for the difference between the two population proportions.

Sample Size for Estimating p1 – p2
The sample size required to obtain a (1 – α)•100% confidence interval with a margin of error, E, is given by rounded up to the next integer, if prior estimates of p1 and p2, , are available. If prior estimates of p1 and p2 are unavailable, the sample size is rounded up to the next integer. The margin of error should always be expressed as a decimal when using Formulas (4) and (5).

B. she does not use any prior estimates.
Parallel Example 5: Determining Sample Size A doctor wants to estimate the difference in the proportion of year old mothers that received prenatal care and the proportion of year old mothers that received prenatal care. What sample size should be obtained if she wished the estimate to be within 2 percentage points with 95% confidence assuming: A. she uses the results of the National Vital Statistics Report results in which 98% of the year old mothers received prenatal care and 99.2% of year old mothers received prenatal care. B. she does not use any prior estimates.

Solution A. We have E = 0.02 and zα/2 = z0.025 = 1.96. Letting
The doctor must sample 265 randomly selected year old mothers and 265 randomly selected year old mothers.

Solution B) Without prior estimates of p1 and p2, the sample size is
The doctor must sample 4802 randomly selected year old mothers and 4802 randomly selected year old mothers. Note that having prior estimates of p1 and p2 reduces the number of mothers that need to be surveyed.

Parallel Example 5: Determining Sample Size
A nutritionist wants to estimate the difference between the proportion of males and females who consume the USDA’s recommended daily intake of calcium. What sample size should be obtained if she wishes the estimate to be within 3 percentage points with 95% confidence, assuming that A. She uses the results of the USDA’s Diet and Health Knowledge Survey, according to which 51.1% males and 75.2% of females consume the USDA’s recommended daily intake of calcium, and B. She does not use any prior estimates?

Inference about Two Means: Dependent Samples
Section 11.2 Inference about Two Means: Dependent Samples

Testing Hypotheses Regarding the Difference
of Two Means Using a Matched-Pairs Design To test hypotheses regarding the mean difference of data obtained form a dependent sample (matched- pairs data), use the following steps. provided that the sample is obtained using simple random sampling, the sample data are matched pairs, the differences are normally distributed with no outliers or the sample size, n, is large (n ≥ 30), the sampled values are independent

Step 1: Determine the null and alternative. hypotheses
Step 1: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways, where μd is the population mean difference of the matched-pairs data.

Classical Approach Step 2: Compute the test statistic which approximately follows Student’s t-distribution with n – 1 degrees of freedom. The values of and sd are the mean and standard deviation of the differenced data.

Classical Approach Step 3: Compare the critical value with the test statistic: Step 4: State conclusion

Parallel Example 2: Testing a Claim Regarding Matched-Pairs Data
The following data represent the cost of a one night stay in Hampton Inn Hotels and La Quinta Inn Hotels for a random sample of 10 cities. Test the claim that Hampton Inn Hotels are priced differently than La Quinta Hotels at the α = level of significance. Note a boxplot and normal probability plot indicate approximately normal with no outliers.

City Hampton Inn La Quinta Dallas 129 105 Tampa Bay 149 96 St. Louis 49 Seattle 189 San Diego 109 119 Chicago 160 89 New Orleans 72 Phoenix 59 Atlanta 90 Orlando 69

Solution This is a matched-pairs design since the hotel prices come from the same ten cities. To test the hypothesis, we first compute the differences and then verify that the differences come from a population that is approximately normally distributed with no outliers because the sample size is small. The differences (Hampton - La Quinta) are: – with = 51.4 and sd =

Solution Step 1: We want to determine if the prices differ:
H0: μd = 0 versus H1: μd ≠ 0 The level of significance is α = 0.05. Step 2: The test statistic is

This is a two-tailed test so the critical values at the α = 0.05 level of significance with n – 1 = 10 – 1 = 9 degrees of freedom are –t0.025 = –2.262 and t0.025 =

Step 3: Since the test statistic, t0 = 5.27 is greater than the critical value t.025 = 2.262, we reject the null hypothesis. Step 4: There is sufficient evidence to conclude that Hampton Inn hotels and La Quinta hotels are priced differently at the α = 0.05 level of significance.

Example Some professor measured the time (in seconds) required to catch a falling meter stick for 12 randomly selected students’ dominant hand and nondominant hand. The professor wants to know if the reaction time in an individual’s dominant hand is less than the reaction time in his or her nondominant hand. A coin flip is used to determine whether reaction time is measured using the dominant or nondominant hand first. Conduct the test at the 𝛼=0.05 level of significance. The data is as follows:

Student Dominant Hand Nondominant Hand 1 0.177 0.179 2 0.210 0.202 3 0.186 0.208 4 0.189 0.184 5 0.198 0.215 6 0.194 0.193 7 0.160 8 0.163 9 0.166 0.209 10 0.152 0.164 11 0.190 12 0.172 0.197

Confidence Interval for Matched-Pairs Data
A (1 – α)•100% confidence interval for μd is given by Lower bound: Upper bound: The critical value tα/2 is determined using n – 1 degrees of freedom. The values of and sd are the mean and standard deviation of the differenced data.

Parallel Example 4: Constructing a Confidence Interval for Matched-Pairs Data Construct a 90% confidence interval for the mean difference in price of Hampton Inn versus La Quinta hotel rooms.

Solution We have already verified that the differenced data come from a population that is approximately normal with no outliers. Recall = 51.4 and sd = From Table VI with α = 0.10 and 9 degrees of freedom, we find tα/2 =

Solution Thus, Lower bound = Upper bound = We are 90% confident that the mean difference in hotel room price for Hampton Inn versus La Quinta Inn is between $33.53 and $69.27.

Example Construct a 95% confidence interval for the dominant vs nondominant hand example to estimate the mean difference, 𝜇 𝑑 .

Inference about Two Means: Independent Samples
Section 11.3 Inference about Two Means: Independent Samples

Sampling Distribution of the Difference of Two
Means: Independent Samples with Population Standard Deviations Unknown (Welch’s t) Suppose that a simple random sample of size n1 is taken from a population with unknown mean μ1 and unknown standard deviation σ1. In addition, a simple random sample of size n2 is taken from a population with unknown mean μ2 and unknown standard deviation σ2. If the two populations are normally distributed or the sample sizes are sufficiently large (n1 ≥ 30, n2 ≥ 30) , then

Sampling Distribution of the Difference of Two
Means: Independent Samples with Population Standard Deviations Unknown (Welch’s t) approximately follows Student’s t-distribution with the smaller of n2 – 1 or n1 – 1 degrees of freedom where is the sample mean and s1 is the sample standard deviation from population 1, and is the sample mean and s2 is the sample standard deviation from population 2.

Testing Hypotheses Regarding the Difference of Two Means
To test hypotheses regarding two population means, μ1 and μ2, with unknown population standard deviations, we can use the following steps, provided that: the samples are obtained using simple random sampling; the samples are independent; the populations from which the samples are drawn are normally distributed or the sample sizes are large (n1 ≥ 30, n2 ≥ 30); for each sample, the sample size is no more than 5% of the population size.

Step 1: Determine the null and alternative. hypotheses
Step 1: Determine the null and alternative hypotheses. The hypotheses are structured in one of three ways:

Step 2: Compute the test statistic
which approximately follows Student’s t- distribution.

Step 3: Compare the critical value with the test statistic:
Step 4: State conclusion

Parallel Example 1: Testing Hypotheses Regarding Two Means
A researcher wanted to know whether “state” quarters had a weight that is more than “traditional” quarters. He randomly selected 18 “state” quarters and 16 “traditional” quarters, weighed each of them and obtained the following data.

Test the claim that “state” quarters have a mean weight that is more than “traditional” quarters at the α = 0.05 level of significance. NOTE: A normal probability plot of “state” quarters indicates the population could be normal. A normal probability plot of “traditional” quarters indicates the population could be normal

Solution Step 1: We want to determine whether state quarters weigh more than traditional quarters: H0: μ1 = μ2 versus H1: μ1 > μ2 The level of significance is α = 0.05. Step 2: The test statistic is

This is a right-tailed test with α = Since n1 – 1 = 17 and n2 – 1 = 15, we will use 15 degrees of freedom. The corresponding critical value is t0.05=1.753.

Step 3: Since the test statistic, t0 = 2.53 is greater than the critical value t.05 = 1.753, we reject the null hypothesis. Step 4: There is sufficient evidence at the α = 0.05 level to conclude that the state quarters weigh more than the traditional quarters.

Example In a Spacelab Life Science 2 payload, 14 male rats were sent to space. Upon their return, the red blood cells mass (in millimeters) of the rats was determined. A control group of 14 male rats was held under the same conditions (except for space flight) as the space rats, and their red blood cell mass was also determined when the space rats returned. The project resulted in the data listed in the following table. Does the evidence suggest that the flight animals have a different red blood cell mass from the control animals at the α = 0.05 level of significance?

Flight Control 8.59 7.89 8.65 8.55 8.64 9.79 6.99 8.70 7.43 6.85 8.40 9.14 7.21 7.54 9.66 7.33 6.39 7.00 7.14 8.58 6.87 8.80 7.62 9.88 9.30 8.03 7.44 9.94

Constructing a (1 – α)•100% Confidence Interval for the Difference of Two Means
A simple random sample of size n1 is taken from a population with unknown mean μ1 and unknown standard deviation σ1. Also, a simple random sample of size n2 is taken from a population with unknown mean μ2 and unknown standard deviation σ2. If the two populations are normally distributed or the sample sizes are sufficiently large (n1 ≥ 30 and n2 ≥ 30), a (1 – α)•100% confidence interval about μ1 – μ2 is given by . . .

Constructing a (1 – α)•100% Confidence Interval for the Difference of Two Means
Lower bound: and Upper bound: where tα/2 is computed using the smaller of n1 – 1 or n2 – 1 degrees of freedom or Formula (2).

Parallel Example 3: Constructing a Confidence Interval for the Difference of Two Means Construct a 95% confidence interval about the difference between the population mean weight of a “state” quarter versus the population mean weight of a “traditional” quarter.

Solution We have already verified that the populations are approximately normal and that there are no outliers. Recall = 5.702, s1 = , = and s2 = From Table VI with α = 0.05 and 15 degrees of freedom, we find tα/2 =

Solution Thus, Lower bound = Upper bound =

Solution We are 95% confident that the mean weight of the “state” quarters is between and ounces more than the mean weight of the “traditional” quarters. Since the confidence interval does not contain 0, we conclude that the “state” quarters weigh more than the “traditional” quarters.

Example Construct a 95% confidence interval for 𝜇 1 − 𝜇 2 using the data presented in the rat example.

Inference about Two Population Standard Deviations
Section 11.4 Inference about Two Population Standard Deviations

Requirements for Testing Claims Regarding Two Population Standard Deviations
1. The samples are independent simple random samples or from a completely randomized design. 2. The populations from which the samples are drawn are normally distributed.

Notation Used When Comparing Two Population Standard Deviations
: Variance for population 1 : Variance for population 2 : Sample variance for population 1 : Sample variance for population 2 n1 : Sample size for population 1 n2 : Sample size for population 2

Fisher's F-distribution
If and and are sample variances from independent simple random samples of size n1 and n2, respectively, drawn from normal populations, then follows the F-distribution with n1 – 1 degrees of freedom in the numerator and n2 – 1 degrees of freedom in the denominator.

Characteristics of the F-distribution
1. It is not symmetric. The F-distribution is skewed right. 2. The shape of the F-distribution depends upon the degrees of freedom in the numerator and denominator. This is similar to the χ2 distribution and Student’s t-distribution, whose shape depends upon their degrees of freedom. 3. The total area under the curve is 1. 4. The values of F are always greater than or equal to zero.

is the critical F with n1 – 1 degrees of freedom in the numerator and n2 – 1 degrees of freedom in the denominator and an area of α to the right of the critical F.

To determine the critical value that has an area of α to the left, use the following property:

Find the critical F-value:
Parallel Example 1: Finding Critical Values for the F-distribution Find the critical F-value: for a right-tailed test with α = 0.1, degrees of freedom in the numerator = 8 and degrees of freedom in the denominator = 4. for a two-tailed test with α = 0.05, degrees of freedom in the numerator = 20 and degrees of freedom in the denominator = 15.

Solution a) F0.1,8,4 = 3.95 b) F.025,20,15 = 2.76

Find the critical F-value:
Parallel Example 1: Finding Critical Values for the F-distribution Find the critical F-value: for a right-tailed test with α = 0.05, degrees of freedom in the numerator = 10, and degrees of freedom in the denominator = 7. for a two-tailed test with α = 0.05, degrees of freedom in the numerator = 15 and degrees of freedom in the denominator = 20.

Test Hypotheses Regarding Two Population Standard Deviations
To test hypotheses regarding two population standard deviations, σ1 and σ2, we can use the following steps, provided that the samples are obtained using simple random sampling. the sample data are independent. the populations from which the samples are drawn are normally distributed.

Two-Tailed Left-Tailed Right-Tailed
Step 1: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways: Two-Tailed Left-Tailed Right-Tailed H0: σ1 = σ2 H1: σ1 ≠ σ2 H1: σ1 < σ2 H1: σ1 > σ2 Note: σ1 is the population standard deviation for population 1 and σ2 is the population standard deviation for population 2.

Step 2: Compute the test statistic
which follows Fisher’s F-distribution with n1 – 1 degrees of freedom in the numerator and n2 – 1 degrees of freedom in the denominator.

Classical Approach Use Table IX to determine the critical value(s) using n1 – 1 degrees of freedom in the numerator and n2 – 1 degrees of freedom in the denominator. The shaded regions in the graphs on the following slides represent the critical regions.

Step 3: Compare the critical value with the test statistic:
Classical Approach Step 3: Compare the critical value with the test statistic: Two-Tailed Left-Tailed Right-Tailed If or , reject the null hypothesis. If , If , Step 4: State conclusion

Parallel Example 2: Testing Hypotheses Regarding Two
Parallel Example 2: Testing Hypotheses Regarding Two Population Standard Deviations A researcher wanted to know whether “state” quarters had a standard deviation weight that is less than “traditional” quarters. He randomly selected 18 “state” quarters and 16 “traditional” quarters, weighed each of them and obtained the data on the next slide. A normal probability plot indicates that the sample data could come from a population that is normal. Test the researcher’s claim at the α = 0.05 level of significance.

Solution Step 1: The researcher wants to know if “state” quarters have a standard deviation weight that is less than “traditional” quarters. Thus H0: σ1 = σ2 versus H1: σ1 < σ2 This is a left-tailed test. The level of significance is α = 0.05.

Solution Step 2: The standard deviation of “state” quarters was found to be and the standard deviation of “traditional” quarters was found to be The test statistic is then

Since this is a left-tailed test, we determine the critical value at the 1 – α = 1 – 0.05 = 0.95 level of significance with n1 – 1=18 – 1=17 degrees of freedom in the numerator and n2 – 1 = 16 – 1 = 15 degrees of freedom in the denominator. Thus, Note: we used the table value F0.05,15,15 for the above calculation since this is the closest to the required degrees of freedom available from Table IX.

Step 3: Since the test statistic F0= 0.52 is greater than the critical value F0.95,17,15=0.42, we fail to reject the null hypothesis.

Solution Step 4: There is not enough evidence to conclude that the standard deviation of weight is less for “state” quarters than it is for “traditional” quarters at the α = 0.05 level of significance.

Example An investor believes that Cisco Systems is a more volatile stock than General Electric. The volatility of a stock is measured by the standard deviation rate of return on the stock. The data in the following table represent the monthly rate of return between 1990 and 2014 for 10 randomly selected months for Cisco Systems stock and 14 randomly selected months for General Electric stock. Does the evidence suggest the Cisco Systems stock is more volatile than General Electric stock at the α = 0.05 level of significance.

Montly Return on Cisco Stock %
Monthly Return on General Electric Stock % 1.93 4.87 1.15 9.00 31.11 5.18 -1.92 3.23 -7.60 8.91 -0.55 -5.19 9.72 23.13 5.29 -3.26 11.64 -5.56 4.68 -0.60 -2.32 0.98 -5.88 4.80

Example In a Spacelab Life Science 2 payload, 14 male rats were sent to space. Upon their return, the red blood cells mass (in millimeters) of the rats was determined. A control group of 14 male rats was held under the same conditions (except for space flight) as the space rats, and their red blood cell mass was also determined when the space rats returned. The project resulted in the data listed in the following table. Is the standard deviation of the red blood cell mass in the flight animals different from the standard deviation of the red blood cell mass in the control animals at the α = 0.05 level of significance.

Flight Control 8.59 7.89 8.65 8.55 8.64 9.79 6.99 8.70 7.43 6.85 8.40 9.14 7.21 7.54 9.66 7.33 6.39 7.00 7.14 8.58 6.87 8.80 7.62 9.88 9.30 8.03 7.44 9.94

Putting It Together: Which Method Do I Use?
Section 11.5 Putting It Together: Which Method Do I Use?

Proportion, p Independent samples: Provided for each sample and the
sample size is no more than 5% of the population size, use the normal distribution with where

σ or σ2 Provided the data are normally distributed, use the
F-distribution with

Mean, μ Independent samples:
Provided each sample size is greater than 30 or each population is normally distributed, use Student’s t-distribution

Mean, μ Dependent samples:
Provided each sample size is greater than 30 or the differences come from a population that is normally distributed, use Student’s t-distribution with n – 1 degrees of freedom with

Example In a study, it found states that distribute chocolate to students prior to teacher evaluations increases results. The authors of the study divided three sections of a course taught by the same instructor into two groups. Fifty of the students were given chocolate by an individual not associate with the course and 50 of the students were not given chocolate. The mean score from the students who received chocolate was 4.2, while the mean score for the non-chocolate group was Suppose the sample standard deviation for both groups was Does chocolate appear to improve teacher evaluations? Use α = 0.1 level of significance.

Example In one particular question on the TIMS exam, a random sample of 400 non-Kumon students resulted in 73 getting the correct answer. For the same question, a random sample of 400 Kumon students resulted in getting the correct answer. Perform the appropriate test to test Kumon’s claims that their students are better. Are there any confounding factors regarding the claims?

Example Do wet suits allow a swimmer to swim faster? Researchers measured the speed in meters per second of swimmers in both with and without a wetsuit. The results of the study are on the next slide. Conduct the appropriate test to determine whether the data suggest that wet suits allow a swimmer to swim faster. Be sure to check requirements for test. Use α = 0.05 level of significance. If you reject null hypothesis, estimate the average difference in speed of the swimmer with the wet suit with 95% confidence.

Swimmer Without wetsuit With wetsuit 1 1.49 1.57 2 1.37 1.47 3 1.35 1.42 4 1.27 5 1.12 1.22 6 1.64 1.75 7 1.59 8 1.52 9 1.50 1.56 10 1.45 1.53 11 1.44 12 1.41 1.51

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