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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6 0

2 years ago

What is 65% of 80 ??????????????

bazaltina [42]

Answer:

Step-by-step explanation:

65% of 80 is

65*80/100=52

4 0