(1 point) Hourly wages at Manufacturing Plant A very from worker to worker, with
ID: 3354718 • Letter: #
Question
(1 point) Hourly wages at Manufacturing Plant A very from worker to worker, with an average hourly wage of $14.47, and a standard deviation of $2.45 per hour, or PlantA-2.45. In an attempt to attract new employees, a newly constructed Manufacturing Plant B claims that its employees will be paid, on average, more than than employees at Manufacturing Plant A. Assume that the standard deviation in hourly wages at Plant B is the same as at Plant A, or PlamB = 2.45 (a) Choose the correct statistical hypotheses A. Ho : PlantB = 14.47 HA·WplantB 14.47 B. Ho . XPlantB = 14.47 HA-XplantB 14.47 C. Ho : XPlantB > 14.47 HA : XPlaniB 14.47 A : HPlantB 14.47 0 : Plant? 14.47 H E. Ho : PlanB 14.47 F. Ho·Anant,-14.47 H (b) Statistical testing of the null hypothesis is to be carried out by randomly selecting 37 employees at Plant B. If the mean/average hourly wage of these 37 employees is greater than $14.69, then there is enough statistical evidence to indicate that the mean hourly wage of employees at Plant B is greater than the mean hourly wage of employees at Plant A. what level of was used here? Enter your answer to at least three decimal places (c) The average of the sample of n- 37 workers was found to be X - 14.59. What decision can you make from this sample? A. Employees at Plant B do earn more on average compared to employees at Plant A B. Employees at Plant B do not earn more on average compared to employees at Plant A. C. The sample size is too small to conduct a statistical test, so a decision cannot be made D. The hourly wages at Plant B need to be Normally distributed to do the test (d) If the average hourly wage of all workers at Plant B is $14.57, what is the probability of concluding using the decision criteria outlined in part (b) that the mean hourly wage of all employees at Plant B is not greater than the mean hourly wage of all employees at Plant A? Enter vour answer to at least three decimals Answer =Explanation / Answer
Solution:-
a) State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: < 14.47
Alternative hypothesis: > 14.47
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.
b) 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 = 0.403
DF = n - 1
D.F = 36
t = (x - ) / SE
t = 0.55
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 0.55.
Thus the P-value in this analysis is 0.293.
Interpret results. Since the P-value (0.293) is greater than the significance level (0.05), we cannot reject the null hypothesis.
c) (B) The employees at plant B do not earn more on average compared to employees at plant A.
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