Question

Instructions Using the Example Independent Samples t-test on page 346 of the book, conduct a 4-step hypothesis test: a. Enter the data into SPSS in the format required to run an independent samples t-test b. Clearly state each of the steps (make sure to use the appropriate notation for the null and alternative hypotheses. (please write up the four step hypothesis test and just screen shot how to do the work on SPSS) Example Independent-Samples t-Test Now the health psychologist wants to compare the calorie estimates of people who regularly eat junk food with the estimates of people who rarely eat junk food. He believes the difference could come out in either direction so he decides to conduct a two-tailed test. He collects data from a sample of eight participants who eat junk food regularly and seven participants who rarely eat junk food. The data are as follows: Junk food eaters: 180, 220, 150, 85, 200, 170, 150, 190 Non-junk food eaters: 200, 240, 190, 175, 200, 300, 240 The mean for the non-junk food eaters is 220.71 with a standard deviation of 41.23. The mean for the junk food eaters is 168.12 with a standard deviation of 42.66. He can now compute his t score as follows: 220.71–168.12 = 2.42 41.232 42.662

Answer 1

The step-by-step instructions obtain two-sample independent t-test using SPSS17: → Load the data into the spss17 spreadsheet. → Put the name of the data set as Responses by clicking in the variable view tab → scale. → Put the name of the data set as factor by clicking in the variable view tab scale. → Click Analyze → Compare Means → Independent sample t-test. → Select the variable Response into the test variables box Click Define groups and enter the specified values group1:1 and group2:2 in the box of group variables. → Click ok. The Two-Sample independent T-test output using SPSS17:

T-Test [DataSeto] Group Statistics N Response factor 1.00 2.00 Mean 168.1250 220.7143 Std. Deviation 41.22564 42.66425 Std. Error Mean 14.57547 16.12557 7 Independent Samples Test Levene's Test for Equality of Variances t-test for Equality of Means 95% Confidence Interval of the Difference Sic. df 13 Mean Difference -52.58929 Siq. (2-tailed) .031 Std. Error Difference 21.68311 767 Lower -99.43280 Response 092 Upper -5.74577 -2.425 Equal variances assumed Equal variances not assumed -2.419 12.600 .031 -52.58929 21.73656 -99.70027 -5.47830

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Let 4 and 4 be the population means and let Xand xz be the means of the respective samples. The summary statistics: Sample 1:24 = 8, 74 = 168.1, s = 41.2 Sample2:n, = 7, X2 = 220.7, s2 = 42.7. Step 1: State the null and alternative hypotheses. The null and alternative hypotheses are, respectively We need to use two-tailed test. H,4 = 4 (There is no significant difference between the population means) versus H,:4, 74, (There is significant difference between the population means) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Step 2: Select the distribution to use. Here, the two samples are independent, , and are unknown but equal, and the sample sizes are small but both populations are normally distributed. Hence, all conditions are satisfied. Consequently, we will use the t -distribution. Step3: Determine the rejection and nonrejection regions. The + sign in the alternative hypothesis indicates that the test is two-tailed. The significance level is 0.05. Hence, the area in two-tail distribution is a = 0.05. The formula for calculating the degrees of freedom is df = m +-2 = 8+7-2 =15-2 = 13 df = 131 The degrees of freedom is df = 13) ---- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - The critical values of t for df = 13 and 0.05 area in right-tail of the t - distribution curve are -2.160369. The critical values of t for df = 13 and 0.05 area in left-tail of the t -distribution curve are 2.160369 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Step 4: Calculate the value of the t-test statistic. The value of the test statistic is t for x7 - X, is computed as follows. The formula for calculating the pooled standard deviation is (n-1) s +in-1) SV 2+1, -2 (8-1)1699.554+(7-1)1820.238 V 8+7-2 (7)1699.554 +(6)1820.238 15-2 11896.88+10921.43 V 13 22818.3 V 13 = V1755.254 = 41.89575 8, = 41.89575 ----------------------- Step 5: The value of the standard error of the difference between the means 1 1 Y , = 41.89575 + = 41.8957570.125+0.142857 = 41.895750.267857 = 41.89575(0.517549) = 21.68311 19 - = 21.68311 ---------------------- - is $,- = + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - The formula for calculating the t-test statistic is (7. – )-(4-11) (168.125 - 220.7143)-(0) 21.68311 _-52.5893 21.68311 =-2.42536 t = -2.42 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

The P-value approach: P-value = 2P(t >to) =2(1-P(t<n)) =2(1-P(t <2.42)) =2(1-(= tdist (2.42,13,1))) =2(1-0.984547) =2(0.015453) 0.030905 P-value = 0.030905 <a = 0.05) We do reject the null hypothesis Ho. ------------------------- Conclusion : Since, the test statistic is kl=2.42 >400.= 2.160369 We reject the null hypothesis H. There is sufficient evidence to conclude that the difference between the population means of the junk food eaters and non junk food eaters is significant at the 5% level of significance.