Question

The average costs for two types of advertising were compared using the following data. Ten weeks of costs were recorded for each type of advertising. Use this information to answer the following questions. Advertising Type One Advertising Type Two n1=10 n2=10 x1 = $590 x2 = $575 s12 = $1,600 s22 = $2,400

14. What is the point estimate for the mean of the cost of Type One Advertising? 15. What is the estimated standard error for the point estimate for the mean of the cost of Type One Advertising? Round to two digits past the decimal. 16. What is the point estimate for the mean of the cost of Type Two Advertising? 17. What is the estimated standard error for the point estimate for the mean of the cost of Type Two Advertising? Round to two digits past the decimal. 18. What is the value of the test statistic to test that the mean of population one is equal to 570? 19. What is the numerical value of the point estimate for the difference between the average costs for these two types of advertising? 20. What is the pooled variance estimate based on these two samples? 21. What is the degree of freedom associated with the pooled variance estimate above? 22. What is the estimated standard error for the point estimate for the difference between the means, if the population variances are assumed equal? 23. What is the estimated standard error for the point estimate for the difference between the means, if the population variances are not assumed equal? Is this equal to the answer in the question just above? If so, why are these two equal? 24. What is the appropriate alternative hypothesis if the research question is “Do the data indicate that the average sales for advertising type one is more than the average sales for advertising type two?” 25. What is the numeric value of the test statistic that would be used to attempt to support the alternative hypothesis described in the above question? 26. What is the appropriate alternative hypothesis if the research question is “Do the data indicate that the average sales for advertising type one is greater than 10 dollars more than the average sales for advertising type two?” 27. What is the numeric value of the test statistic that would be used to attempt to support the alternative hypothesis described in the above question? 28. If the sales averages for the two advertising types are equal, what is the name of the distribution of the test statistic, if the standard error is based on the variance estimate that is used when the population variances are assumed equal? 29. In this situation the null hypothesis would be rejected at the 1% significance level if the observed test statistic value is more than what value? 30. dIf the p-value of this hypothesis test is 0.009 would one conclude that the average sales for advertising type one is more than the average sales for advertising type two at the 5% significance level? Answer with a sentence for the conclusion.

Answer 1

14. The point estimate for the mean of the cost of Type One Advertising is $590

15.The estimated standard error for the point estimate for the mean of the cost of Type One Advertising is sqrt(s1^2/n1)

sqrt(1600/10)= 40/sqrt(10)= 40/3.16= 12.66

16.The point estimate for the mean of the cost of Type Two Advertising is $575

17.The estimated standard error for the point estimate for the mean of the cost of Type Two Advertising is sqrt(2400/10)= 15.49

18. The value of the test statistic to test that the mean of population one is equal to 570 is

t= xbar-mean/s/sqrt(n)

t= 590-570/12.66

t= 20/12.66

t= 1.58

19.The numerical value of the point estimate for the difference between the average costs for these two types of advertising is $(590-575) = $15

20.The following formula is used to compute the pooled variance:

sp^2​=(n1​−1)s1^2​+(n2​−1)s2^2​​/(n1​+n2​−2)

Plugging in the corresponding values in the formula above, we get that

sp^2​=(10−1)(40)^2+(10−1)(48.99)^2/(​10+10−2) =2000.01

sp= sqrt(2000.01)= 44.72

21.The degree of freedom associated with the pooled variance estimate above is n1+n2-2= 10+10-2= 18

22. The estimated standard error for the point estimate for the difference between the means, if the population variances are assumed equal

sp*sqrt(1/n1+1/n2)

44.72*sqrt(1/10+1/10)

44.72*0.45

20.12

NOTE: As we are allowed to solve first four questions. I have done with steps first six. Please re post rest with this information (Advertising Type One Advertising Type Two n1=10 n2=10 x1 = $590 x2 = $575 s12 = $1,600 s22 = $2,400)