Vail Resorts tracks the proportion of seasonal employees who are rehired each se
ID: 3233068 • Letter: V
Question
Vail Resorts tracks the proportion of seasonal employees who are rehired each season. Rehiring a seasonal employee is beneficial in many ways, including lowering the costs incurred during the hiring process such as training costs. A random sample of 840 full-time and 438 part-time seasonal employees from 2009 showed that 446 full-time employees were rehired compared with 213 part-time employees.
39.
value:
10.00 points
Required information
Is there a significant difference in the proportion of rehires between the full-time and part-time seasonal employees? (Use = 0.10)
Choose the appropriate hypotheses. Assume F is the proportion of full-time employees and P is the proportion of part-time employees.
Specify the decision rule. (Negative value should be indicated by a minus sign. Round your answers to 3 decimal places.)
Find the test statistic zcalc. (Do not round the intermediate calculations. Round your answer to 3 decimal places.)
References
eBook & Resources
WorksheetLearning Objective: 10-05 Perform a test to compare two proportions using z.
Difficulty: 2-MediumLearning Objective: 10-07 Use Excel to determine p-values for two-sample tests using z or t.
Check my work
40.
value:
10.00 points
Required information
References
eBook & Resources
WorksheetLearning Objective: 10-05 Perform a test to compare two proportions using z.
Difficulty: 2-MediumLearning Objective: 10-07 Use Excel to determine p-values for two-sample tests using z or t.
Check my work
Vail Resorts tracks the proportion of seasonal employees who are rehired each season. Rehiring a seasonal employee is beneficial in many ways, including lowering the costs incurred during the hiring process such as training costs. A random sample of 840 full-time and 438 part-time seasonal employees from 2009 showed that 446 full-time employees were rehired compared with 213 part-time employees.
Explanation / Answer
Solution:-
1) State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P1 = P2
Alternative hypothesis: P1 P2
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion from population 1 is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.10. The test method is a two-proportion z-test.
Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).
2) zcritical = + 1.645
We will reject the null hypothesis if z value of test lie outside the range - 1.645 < z < 1.645.
p = (p1 * n1 + p2 * n2) / (n1 + n2)
p = 0.51565
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.02945
3) z = (p1 - p2) / SE
z = 1.52
where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.
4) Since z value does not lie in the rejection region hence we have to accept the null hypothesis.
5) We cancannot conclude that there is a significant difference in the proportion of rehires between the full-time and part-time seasonal employees.
Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 1.52 and more than 1.52.
We use the Normal Distribution Calculator to find P value = 0.1286
Interpret results. Since the P-value (0.1286) is greater than the significance level (0.10), we cannot reject the null hypothesis.
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