Question

A Ski resort 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 833 full-time and 386 part-time seasonal employees from 2009 showed that 434 full-time employees were rehired compared with 189 part-time employees (a) Is there a significant difference in the proportion of rehires between the full-time and part-time seasonal employees? (Use α = 0.10) (a-1) Choose the appropriate hypotheses. Assume TTF is the proportion of full-time employees and πrho is the proportion of part-time employees (a-2) Specify the decision rule. (A negative value should be indicated by a minus sign. Round your answers to 3 decimal places.)
(a-3) Find the test statistic Zcalc (Do not round the intermediate calculations. Round your answer to 3 decimal places.)

Answer 1

The test statistic z will be equal to -0.946 and it shows that there is no significant difference in the proportion of rehires between full time and part time.

Given sample sizes of 833 and 386 and result of samples 434 and 189.

$n_{1} =833, n_{2}=386$

Proportion  of full time=434/833=0.52

Proportion of part time=189/386=0.49.

Difference in proportion =0.52-0.49

$H_{0}:$TTF- i∈ rho=0

$H_{1}:$ TTF+i∈ rho≠0.

Mean of difference=0.03

Z=(X-μ)/σ

σ=$\sqrt{0.4517/447}$

=0.0317

σ=0.0317

z=(0-0.03)/0.0317

=-0.03/0.0317

=-0.317

p value will be =0.1736.

Because p value is greater than 0.01 so we will accept the null hypothesis which shows that there is no significant difference in the proportions.

Hence there is no significant difference in the proportion of rehires between full time and part time.

Learn more about z test at brainly.com/question/14453510

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