34. Increasing numbers of businesses are offering child-care benefits for their

Question

34. Increasing numbers of businesses are offering child-care benefits for their workers. However, one union claims that more than 80% of firms in the manufacturing sector still do not offer any child-care benefits to their workers. A random sample of 360 manufacturing firms is selected and asked if they offer child-care benefits. Suppose the P value for this test was reported to be p-0.1149. State the conclusion of interest to the union. Use -0.05. (4pts) st was reported to be p 0.1149. State the conclusion of interest to the

Explanation / Answer

Solution:-

34)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P < 0.80

Alternative hypothesis: P > 0.80

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Thus the P-value in this analysis is 0.1149

Interpret results. Since the P-value (0.1149) is greater than the significance level (0.05), we cannot reject the null hypothesis.

From the above test we do not have sufficient evidenc ein the favor of the claim that more than 80% of the firms in the manufacturing sector still do not offer any child care benefits to their workers.

21)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: > 2.2

Alternative hypothesis: < 2.2

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.1789

DF = n - 1 = 20 - 1

D.F = 19

t = (x - ) / SE

t = - 1.118

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of - 1.118. We use the t Distribution Calculator to find P(t < - 1.118) = 0.1388

Thus the P-value in this analysis is 0.1388

Interpret results. Since the P-value (0.1388) is greater than the significance level (0.05), we cannot reject the null hypothesis.

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