Question

5. (30 points) A food company is developing a new granola bar, and its market analysts are currently working on preliminary studies of the packaging design. To help with a marketing strategy, the company was first interested in whether the appeal of the packaging design for the new product (and hence the appeal of the product itself) was related to a person's gender. There were 100 male and 100 female volunteers available for purposes of evaluation. Both males and females rated the design on a scale of 1 to 10, 1 being "very unappealing” and 10 being "very appealing”. The mean rating for males was 7.4 with a standard deviation 1.5. The mean rating for females was 8.0 with a standard deviation 2.0. Assume our samples can be regarded as random samples from the populations of males and females. 1) (10 points) Is there good evidence that the mean rating for the population of males is different from that for the population of females? State the Ho and Ha and carry out an appropriate test statistic. Give the P-value and state your conclusion using significance level a = 0.05. 2) (5 points) Give a 95% confidence interval for the difference in the mean ratings between males and females.

Answer 1

The provided sample means are shown below: X1 = 7.4 X2 = 8 Also, the provided sample standard deviations are: Si = 1.5 S2 = 2 and the sample sizes are ni = 100 and n2 = 100. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = 42 Ha: Mi * u2 This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the degrees of freedom are df = 198. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal: = 1.972, for Hence, it is found that the critical value for this two-tailed test is tc a = 0.05 and df = 198. The rejection region for this two-tailed test is R= {t:1t| > 1.972}. .. .(3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: t= X1 - 12 (n1-1) +(n2-1) 1 ni+n2-2 ni + 7.4 - 8 2.4 (100-1)1.52+(100-1)22 100+100-2 (100 + 100) (4) Decision about the null hypothesis Since it is observed that \t\ = 2.4 > tc = 1.972, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0173, and since p= 0.0173< 0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean mi is different than 42, at the 0.05 significance level. Confidence Interval The 95% confidence interval is - 1.093 < M1 – M2 < -0.107.

The provided sample mean is X = 7.4 and the sample standard deviation is s = 1.5, and the sample size is n 100. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u = = 5 Ha: > 5 This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a right-tailed test is te = 1.66. The rejection region for this right-tailed test is R=t:t > 1.66 (3) Test Statistics The t-statistic is computed as follows: X — по t 7.4 - 5 1.5/V100 16 s/vn (4) Decision about the null hypothesis Since it is observed that t = 16 > t = 1.66, it is then concluded that the null hypothesis is rejected. Using the P-value approach The p-value is p = 0, and since p= 0 < 0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean u is greater than 5, at the 0.05 significance level. Confidence Interval The 95% confidence interval is 7.102 < p < 7.698.

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