Do shoppers at the mall spend the same amount of money on average the day after Thanksgiving comparedto the day after Christmas? The 40 randomly surveyed shoppers on the day after Thanksgiving spent anaverage of S146. Their standard deviation was $37. The 48 randomly surveyed shoppers on the day afterChristmas spent an average of $166. Their standard deviation was $25. What can be concluded at the a =0.01 level of significance?For this study, we should use Select an answera. The null and alternative hypotheses would be:Ho: Select an answer VSelect an answerSelect an answer vH:Select an answer vSelect an answerVSelect an answer Vb. The test statistic ?v(please show your answer to 3 decimal places.)c. The p-value =(Please show your answer to 4 decimal places.)d. The p-value is 7vae. Based on this, we should Select an answer v the null hypothesis.f. Thus, the final conclusion is thatO The results are statistically significant at a= 0.01, so there is sufficient evidence to concludethat the mean expenditure for the 40 day after Thanksgiving shoppers that were observed is adifferent amount of money compared to the mean expenditure for the 48 day after Christmasshoppers that were observed.O The results are statistically significant at a = 0.01, so there is sufficient evidence to concludethat the population mean amount of money that day after Thanksgiving shoppers spend is adifferent amount of money compared to the population mean amount of money that day afterChristmas shoppers spend.O The results are statistically insignificant at a = 0.01, so there is insufficient evidence toconclude that the population mean amount of money that day after Thanksgiving shoppersspend is a different amount of money compared to the population mean amount of money thatday after Christmas shoppers spend.O The results are statistically insignificant at a = 0.01, so there is statistically significantevidence to conclude that the population mean amount of money that day after Thanksgivingshoppers spend is equal to the population mean amount of money that day after Christmasshoppers spend.

The sample size on the day after Thanksgiving is 40, with average of $146, and standard deviation was $37. The sample size on the day after Christmas spent is 48 with an average of $166, and the standard deviation was $25.It is clear that the population standard deviation is not known and sample sizes are greater than 30 which are large indicating that t-test for the difference between two independent population means is used for this study. Therefore, the t-test for the difference between two independent population means is used for this study.a. Let μ1 be the population mean amount of money spent day after thanksgiving, and μ2 be the population mean amount of money spent day after Christmas. The claim is to check whether the mall spend the same amount of money on average the day after Thanksgiving compared to the day after Christmas. The hypothesis is, Null hypothesis: H0:μ1=μ2 Alternative hypothesis: H1:μ1≠μ2b. The test statistic is, t=x¯1-x¯2s12n1+s22n2=146-16637240+25248=-206.873560=-2.9097≈-2.910 Therefore, the test statistic is -2.910.c. The degrees of freedom is, df=s12n1+s22n22s12n12n1-1+s22n22n2-1=37240+25248237240240-1+25248248-1≈66 The hypothesis is two tailed. The probability of t not equal to 2.910 with 66 degrees of freedom can be obtained using the excel formula “=T.DIST.2T (2.910, 66)”. The p-value is 0.0049. Therefore, the p-value is 0.0049.d , e, f. Decision rule: If p-value≤α, then reject the null hypothesis. Conclusion: The p-value (0.0049) is less than the level of significance (0.01). Based on decision rule, reject the null hypothesis. Thus, the results are statistically significant at α=0.01, so there is sufficient evidence to conclude that the population mean amount of money that day after Thanksgiving shoppers spend is a different amount of money compared to the population mean amount of money that day after Christmas shoppers spend.