Check My Work The Intemal Revenue Service (IRS) provides atoll-free help line fo
ID: 3351724 • Letter: C
Question
Check My Work The Intemal Revenue Service (IRS) provides atoll-free help line for taxpayers to call in and get answers lo questions as they prepare theirtax returns. In recent yeans, the IRS has been inundated with taxpayer calls and has redesigned its phone service as well as posting answers to frequently asked questions on its website (The Cncnna Enquirer, January 7, 2030).According to a report by a taxpayer advocate, calers using the new system can expect to wat on hold for an unreesonably long time of s 0 minutes before g able to talk to an IRS employee. Suppose yo·wet a sample of50calers tr ne-phom srweha·been meeneme, the onge mur, show e mean waiting time of 13 minutes before an IRS employee comes on ine. Based upon data from past years, you decide it is reasonable to assume that the standard deviation of wating times is 11 minutes. Use -0.05. a. State the hypotheses b. what is the pvalue (to 4 decimais)? 0.03803 c. Using a -0.05, can you conchude that the actual mean mating time is signfhcantly ess tha the clam of 15 minutes made by the taxpayer advecate Partially CorreetExplanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: > 15
Alternative hypothesis: < 15
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n)
S.E = 1.556
DF = n - 1
D.F = 49
t = (x - ) / SE
t = - 1.29
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
The observed sample mean produced a t statistic test statistic of - 1.29.
Thus the P-value in this analysis is 0.102.
Interpret results. Since the P-value (0.102) is greater than the significance level (0.05), we cannot reject the null hypothesis.
From the above test we do not have sufficient evidence in the favor of the claim that waiting time is less than 15 minutes.