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 .

Since the p value is higher than the significance level given of [tex]\alpha=0.05[/tex] 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.

Step-by-step 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: [tex]p \leq 0.85[/tex]

Alternative hypothesis: [tex]p >0.85[/tex]

And the statistic for this test is given by:

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

The p value can be calculated with this formula:

[tex] p_v = P(z>z_{calc}) [/tex]

And the p value for this test is given [tex]p_v = 0.1071[/tex]

Since the p value is higher than the significance level given of [tex]\alpha=0.05[/tex] 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.