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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.

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: $p \leq 0.85$

Alternative hypothesis: $p >0.85$

And the statistic for this test is given by:

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

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.

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3 years ago