Question

Last year, the mean dollar spent for online purchases on the AMAZING WEBSITE for the week (7 days) before Labor Day among customers who use their Vista Charge Card was $350. The population standard deviation is not known. Because of increased use of on-line purchasing, Vista’s Vice President of Electronic Marketing believes that purchasing on the AMAZING WEBSITE has changed. He randomly selects 100 customer accounts. The results of the sample found that customers that used Vista charge card spent a mean dollar amount of $295 on purchases with s = $55. (use alpha 1%)

a) State the null and alternative hypothesis in symbols and in words. (3pts)

b) Calculate the expected results for the hypothesis test sampling distribution, assuming the null hypothesis is true. (i.e., name and graph of sampling distribution, mean and standard error) (3pts)

c) Identify the standard distribution that best approximates the sampling distribution. (1pts)

d) Formulate the decision rule(use can use either critical test scores or p-values) (3pts)

e) Determine the statistical results: test statistic, critical test statistic and p-value ( 6pts)

f) Determine the conclusion in terms of the null and alternative hypotheses. Do the sample results indicate that the vice president’s claim is supported at alpha = 1%?

Answer 1

Hypothesized Mean dollar spent = 350

Mean dollar spent from sample, = 295,

and std deviation of dollar spent, s = 55

a) Null hypothesis is that the mean dollar spent is 350, i.e. unchanged. Alternative hypothesis is that the mean dollar spent has changed, i.e. is different from 350.

H₀:   = 350

H_a:   350

b) As the population variance is not known, the sampling distribution is the Student's t-distribution, with a graph like that of the normal distribution but with fatter tails.

Mean of this distribution = 295, and std error of the sample = 55 / sqrt(100) = 5.5

c) The standard distribution that best approximates the sampling distribution, is the Std normal distribution with

mean = 0, and std error = 55 / sqrt(100) = 5.5, i.e. N(0, 5.5)

d) The decision rule is that if the t-statistic of the sample is greater than the critical t-score with d.o.f = 100-1 =99 at the chosen significance level of 1%, we reject the null hypothesis, else fail to reject it at the 1% significance level.

e) Test statistic, t-statistic = [ x̄ - μ ] / [ s / sqrt( n ) ] = (295 - 350) /  5.5 = -10

Critical t-score (2-sided), t_{0.005, 99} = 2.626

And, p-value ~ 0

f) Since, abs(t-statistic) >= t_{0.005, 99} = 10,

We reject the null hypothesis at the 1% significance level in favor of the alternative hypothesis, i.e. the Vice President's claim. Hence, the Vice President's claim is supported at alpha = 1%