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In: Statistics and Probability

Increasing numbers of businesses are offering child-care benefits for their workers. However, one union claims that...

Increasing numbers of businesses are offering child-care benefits for their workers. However, one union claims that more than 85% of firms in the manufacturing sector still do not offer any child-care benefits. A random sample of 460 manufacturing firms is selected, and only 34 of them offer child-care benefits. Specify the rejection region that the union will use when testing at alpha=.10 include the null and alternate hypothesis

Expert Solution

Union claims that more than 85% of firms in the manufacturing sector still do not offer any child-care benefits.

A random sample of 460 manufacturing firms is selected, and only 34 of them offer child-care benefits

Number of manufacturing firms who do not offer any child-care benefits = 460 - 34 = 426

Sample proportion of manufacturing firms who do not offer any child-care benefits = = 426/460 = 0.9261

Sample size = n = 460

Hypothesis :

Right tailed test.

Rejection region :

Given significance level = = 0.1

Critical value for this right tailed test is ,

{ Using Excel, =NORMSINV(0.9) = 1.2816 }

Rejection region = { z : z > 1.2816 }

Test statistic :

Decision about null hypothesis :

It is observed that test statistic (4.571) is greater than critical value (1.2816 )

So reject null hypothesis .

Conclusion :

There is sufficient evidence to support the claim that more than 85% of firms in the manufacturing sector still do not offer any child-care benefits.

orchestra answered 1 year ago

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Increasing numbers of businesses are offering child-care benefits for their workers. However, one union claims that...

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