Question

Do shoppers at the mall spend less money on average the day after Thanksgiving compared to the day after Christmas? The 57 randomly surveyed shoppers on the day after Thanksgiving spent an average of $143. Their standard deviation was $30. The 53 randomly surveyed shoppers on the day after Christmas spent an average of $138. Their standard deviation was $36. What can be concluded at the αα = 0.10 level of significance? Assuming Equal Variances  

For this study, we should use Select an answer t-test for the difference between two dependent population means z-test for a population proportion z-test for the difference between two population proportions t-test for the difference between two independent population means t-test for a population mean

1. The null and alternative hypotheses would be:   
2.   

H0:H0:  Select an answer μ1 p1  Select an answer ≠ ≥ ≤ =  Select an answer p2 μ2  (please enter a decimal)   

H1:H1:  Select an answer μ1 p1  Select an answer = ≠ < >  Select an answer p2 μ2  (Please enter a decimal)

2. The test statistic is a ? z t
3. The test statistic value =  (please show your answer to 3 decimal places.)
4. The p-value =  (Please show your answer to 4 decimal places.)
5. The p-value is ? ≤ >  αα
6. Based on this, we should Select an answer fail to reject reject accept  the null hypothesis.
7. Thus, the conclusion is that ... Select an answer At the 10% significance level, the data provides sufficient evidence to conclude that the population mean amount of money that day after Thanksgiving shoppers spend is less than the population mean amount of money that day after Christmas shoppers spend. At the 10% significance level, the data provides insufficient evidence to conclude that the population mean amount of money that day after Thanksgiving shoppers spend is less than the population mean amount of money that day after Christmas shoppers spend.

Answer 1

The provided sample means are shown below: XL = = 143 X2 = 138 Also, the provided sample standard deviations are: $1 = 30 S2 = 36 and the sample sizes are ni = :57 and n2 = 53. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = 42 Ha: ui M2 This corresponds to a left-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 = 0.10, and the degrees of freedom are df = 108. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal: Hence, it is found that the critical value for this left-tailed test is te = -1.289, for 0.10 and df 108. a = The rejection region for this left-tailed test is R = {t:t< -1.289}. (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: t= X1 - X2 (ni-1)s+(n2-1)s ni+n2-2 + m) ni 143 – 138 0.793 (57-1)302+(53-1)362 57+53-2 (57 +53) (4) Decision about the null hypothesis Since it is observed that t = 0.793 > te = -1.289, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p=0.7854, and since p = 0.7854 > 0.10, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean Mi is less than M2, at the 0.10 significance level.

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