T-Test for Two Independent Samples

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Hypothesis Test: T-Test for Independent Samples When Both σ s Are Unknown In this hypothesis test, we want to assess a claim about means from two independent samples where both population standard deviations are unknown. We will also assume that the population variances are either unknown or unequal. To run this test, we need to use the following test statistic: Test Statistic Used for T-Test for Two Independent Samples Note that because the absolute t-value is used to determine the p-values, the sign (+/-) of t-statistics obtained from hand calculation might differ from the sign obtained through technology. Now let's see an example using summary statistics.The following problem uses hand calculations and then technology to determine the p-value: T-Test for Two Independent Samples https://leocontent.umgc.edu/content/umuc/tus/stat/stat200/2232/t-test-fo... 1 of 12 3/20/24, 11:51 PM
Example: T-Test for Independent Samples (Hand Calculation Using Summary Statistics) A sports physiologist thinks that, on average, biking burns more calories than aerobic dancing. To test his hypothesis, the physiologist randomly selects 12 female bike riders and 15 women who take low-impact aerobics classes. Each participant engaged in her activity for 30 minutes, during with oxygen usage was measured. These measures were then converted into the number of calories each person had burned, and means and standard deviations were calculated. Step 1: Identify the Question/Problem to Solve Using a significance level of 0.05, is the mean number of calories burned greater for women riding bikes than for women taking a low-impact aerobics class? State the null and alternative hypotheses (H and H ). H : μ = μ H : μ > μ In words, the hypotheses are as follows: H : The mean number of calories burned by biking is equal to the mean number of calories burned doing low-impact aerobics. H : The mean number of calories burned by biking is greater than the mean number of calories burned doing low-impact aerobics. Because there is a "greater than" alternative hypothesis, this is a one-tailed test. Step 2: Identify and Collect Data One continuous random variable that is ratio level. The raw data for the problem is contained in the following table: o a o 1 2 a 1 2 o a T-Test for Two Independent Samples https://leocontent.umgc.edu/content/umuc/tus/stat/stat200/2232/t-test-fo... 2 of 12 3/20/24, 11:51 PM
BikingLow Impact Aerobics n 12 15 148 136 s 11 5 Step 3: Select and Apply the Appropriate Method Since we are assessing the difference between two sample means and it meets the following assumptions, the appropriate statistical test is the independent samples t-test, where the sample standard deviations are unknown or assumed to be unequal. Assumptions: • The two independent samples are simple random samples. • The two independent samples are from two distinct populations. • For the two distinct populations, ◦ If the sample sizes are small, the distributions are important (should be normal). ◦ If the sample sizes are large (>30), the distributions are not important (need not be normal). • The population variances are unknown and assumed to be not equal. The test statistic is as follows: T-Test for Two Independent Samples https://leocontent.umgc.edu/content/umuc/tus/stat/stat200/2232/t-test-fo... 3 of 12 3/20/24, 11:51 PM
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Since we are assuming that the population variances are not equal, the Welch– Satterthwaite equation is used to calculate degrees of freedom: We can now calculate the p-value using the p-value from t-score web applet (detailed instructions available here: P-Value from t-score Applet (https:// leocontent.umgc.edu/content/umuc/tus/stat/stat200/2232/p-value-from-t- score-applet.html?ou=935683) ): T-Test for Two Independent Samples https://leocontent.umgc.edu/content/umuc/tus/stat/stat200/2232/t-test-fo... 4 of 12 3/20/24, 11:51 PM
Decide whether or not to reject H The p -value (0.001668) is much smaller than the given α , which is 0.05. Since the p -value is less than the significance level, there is enough evidence to warrant rejection of the null hypothesis that the two means are the same. Step 4: Interpret the Results and Apply to Real Life The average number of calories burned biking is significantly larger than the average number of calories burned doing low-impact aerobics (p<.05). Now let's see how to use the free QuickCals applet to do all the calculations o. T-Test for Two Independent Samples https://leocontent.umgc.edu/content/umuc/tus/stat/stat200/2232/t-test-fo... 5 of 12 3/20/24, 11:51 PM
using all of the same data and assumptions as above: T-Test for Two Independent Samples https://leocontent.umgc.edu/content/umuc/tus/stat/stat200/2232/t-test-fo... 6 of 12 3/20/24, 11:51 PM
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Example: T-Test for Independent Samples Using Summary Statistics in a Free Web Applet Note about output from this applet: The calculated p -value is for a two-tailed test. Since we are doing a one-tailed test, the p- value needs to be divided by 2; therefore, the p -value for this problem is: 0.0035/2 =0.00175. As we can see, the t-value is (rounded to two decimal places) 3.50 with 14 degrees of freedom, with a p -value of 0.00175. Note: Notice how the p-value and degrees of freedom are slightly different than those obtained through using hand calculation. These results are more accurate, since the entire t-test, including degrees of freedom, were calculated using technology. T-Test for Two Independent Samples https://leocontent.umgc.edu/content/umuc/tus/stat/stat200/2232/t-test-fo... 7 of 12 3/20/24, 11:51 PM
Finally, let's see how to calculate an independent samples t-test from raw data using technology: T-Test for Two Independent Samples https://leocontent.umgc.edu/content/umuc/tus/stat/stat200/2232/t-test-fo... 8 of 12 3/20/24, 11:51 PM
Example: t-Test for Independent Samples (Using Raw [Individual Level] Data) Are food products identified as "healthy alternatives" really nutritionally better than their "regular" counterparts? In order to answer this question, a researcher collected data from frozen entrees at a commissary on an American base in Germany. The entrees were first divided by type, regular or healthy alternative, then 22 regular and 20 healthy alternative boxes were randomly selected for the study. Data on the number of milligrams of cholesterol per serving of each entree box was recorded. Step 1: Identify the Question/Problem to Solve Using the .05 significance level, determine whether or not there is a difference in cholesterol between "regular" and "healthy alternative" frozen entrees. State the null and alternative hypotheses (H and H ). H : μ = μ H : μ ≠ μ In words, the hypotheses are: H : The mean number of milligrams of cholesterol in "regular" frozen entrees is equal to the mean number of milligrams of cholesterol in "healthy alternative" frozen entrees. H : The mean number of milligrams of cholesterol in "regular" frozen entrees is unequal to the mean number of milligrams of cholesterol in "healthy alternative" frozen entrees.. Because there is an "unequal" alternative hypothesis, this is a two-tailed test. The problem provides the significance level to use: 0.05. Step 2: Identify and Collect Data o a o 1 2 a 1 2 o a T-Test for Two Independent Samples https://leocontent.umgc.edu/content/umuc/tus/stat/stat200/2232/t-test-fo... 9 of 12 3/20/24, 11:51 PM
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One continuous random variable that is ratio level: milligrams of cholesterol per serving. Here are the summary statistics: RegularHealthy Alternative n 20 22 40.00 31.50 s 20.90 10.84 Step 3: Select and Apply the Appropriate Method The appropriate statistical test is the independent samples t-test, where the sample standard deviations are unknown or assumed to be unequal. Assumptions: • The two independent samples are simple random samples. • The two independent samples are from two distinct populations. • For the two distinct populations, ◦ If the sample sizes are small, the distributions are important (should be normal). ◦ If the sample sizes are large (>30), the distributions are not important (need not be normal). • The population variances are unknown and not assumed to be equal. T-Test for Two Independent Samples https://leocontent.umgc.edu/content/umuc/tus/stat/stat200/2232/t-test-fo... 10 of 12 3/20/24, 11:51 PM
Here are the calculations using the free t-test calculator applet: T-Test for Two Independent Samples https://leocontent.umgc.edu/content/umuc/tus/stat/stat200/2232/t-test-fo... 11 of 12 3/20/24, 11:51 PM
Because the alternative hypothesis is "unequal to," this is a two-tailed test. As we can see from the results, the t value is 1.6302, with 27 degrees of freedom, and a p-value of 0.1147. Decide whether or not to reject H Because the p-value of 0.1147 is more than the significance level of 0.05, we fail to reject the null hypothesis that the mean number of milligrams of cholesterol in "regular" frozen entrees (mean=40.00) is equal to the mean number of milligrams of cholesterol in "healthy alternative" frozen entrees (mean=31.50). Step 4: Interpret the Results and Apply to Real Life There is not enough evidence to show that there is a difference in mean cholesterol levels between regular entrees and healthy alternative entrees (p>. 05). o. T-Test for Two Independent Samples https://leocontent.umgc.edu/content/umuc/tus/stat/stat200/2232/t-test-fo... 12 of 12 3/20/24, 11:51 PM
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