Refer to the following for Questions 10-15. An entrepreneur is considering the p
ID: 2947859 • Letter: R
Question
Refer to the following for Questions 10-15. An entrepreneur is considering the purchase of a coin- operated laundry. The present owner claims that over the past 5 vears, the mean daily revenue was at S700. A sample of 60 days' revenue is taken (sample mean is $676 and the sample standard deviation is S75). The entrepreneur believes this claim may be inflated and the true mean is actually significantly less than $700 in which case he will not buy the laundry. Questions 10-15 EXCEL OUTPUT Data Data ull Hypothesis 700 0.01 700Null Hypothesis 0.01Level of Significance 75Sample Size vel of Significance ulation Standard Deviation ample Size ample Mean andard Error of the Mean Test Statistic 67 60Sample Mean 676Sample Standard Deviation 9.6 9.68 Standard Error of the Mean ees of Freedom Test Statistic Lower-Tail Test Lower-Tail Test Lower Critical Value p-Value 2.326 Lower Critical Value 0.0066p-Value 2.391 0.008 Use the Excel output for Questions 10-15 to test if the data provide evidence (at ?- 01) that the true mean daily revenue is significantly less than $700. 10. The alternative hypothesis will be a) ? #5700 b) ?> $700 c) H $700 d) S676 11. The correct test statistic for this problem is x-700 x-676 x-700 x-676 a) 12. The value of the test statistic for this problem is a) -2.326 b) -2.3912 c) -2.48 d) 2.48 13. The statistical decision rule of the hypothesis test is a) Reject Ho if t-stat> 2.6618 b) Reject Ho if z-stat 2.576 d) Reject Ho if tExplanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: u > 700
Alternative hypothesis: u < 700
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.01. 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 = 9.68
DF = n - 1
D.F = 59
t = (x - u) / SE
t = - 2.48
tcritical = - 2.391
Rejection region is t < - 2.391
where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.
The observed sample mean produced a t statistic test statistic of - 2.48.
Thus the P-value in this analysis is 0.008.
Interpret results. Since the P-value (0.008) is less than the significance level (0.01), we have to reject the null hypothesis.
From the above test we have sufficient evidence in the favor of the claim that true mean is actually significantly less than 700.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.