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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 to their workers. random sample of 330 manufacturing firms is selected and asked if they offer child-care benefits. Suppose the P-value for this test was reported to be p = 0.1071. State the conclusion of interest to the union. Use alpha=0.05 .

User Asifmohd
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Answer:

The p value for this test is given
p_v = 0.1071

Since the p value is higher than the significance level given of
\alpha=0.05 we have enough evidence to FAIL to reject the null hypothesis. And we can say that the true proportion of firms in the manufacturing sector still do not offer any child-care benefits to their workersis is not significantly higher than 0.85 or 85% at 5% of significance.

Explanation:

We define the proportion of interest as p who represent the true proportion of firms in the manufacturing sector still do not offer any child-care benefits to their workers

And we want to anaylze the following system of hypothesis:

Null hypothesis:
p \leq 0.85

Alternative hypothesis:
p >0.85

And the statistic for this test is given by:


z =\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}

The p value can be calculated with this formula:


p_v = P(z>z_(calc))

And the p value for this test is given
p_v = 0.1071

Since the p value is higher than the significance level given of
\alpha=0.05 we have enough evidence to FAIL to reject the null hypothesis. And we can say that the true proportion of firms in the manufacturing sector still do not offer any child-care benefits to their workersis is not significantly higher than 0.85 or 85% at 5% of significance.

User Puffin
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