The average duration of labor from the first contraction to the birth of the baby in women over 35 who have not previously given birth and who did not use any pharmaceuticals is 16 hours. Suppose you have a sample of 29 women who exercise daily, and who have an average duration of labor of 17.8 hours and a sample variance of 77.4 hours. You want to test the hypothesis that women who exercise daily have a different duration of labor than all women. Calculate the t statistic. To do this, you first need to calculate the estimated standard error. The estimated standard error is s M

There is not enough evidence to support the claim that women who exercise daily have a significantly different duration of labor than all women.

Standard error sm = 1.634

Test statistic t = 1.102

P-value = 0.28

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that women who exercise daily have a significantly different duration of labor than all women.

Then, the null and alternative hypothesis are:

The significance level is 0.05.

The sample has a size n=29.

The sample mean is M=17.8.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=√77.4=8.8.

The estimated standard error of the mean is computed using the formula:

Then, we can calculate the t-statistic as:

The degrees of freedom for this sample size are:

This test is a two-tailed test, with 28 degrees of freedom and t=1.102, so the P-value for this test is calculated as (using a t-table):

As the P-value (0.28) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that women who exercise daily have a significantly different duration of labor than all women.

6.9

step-by-step explanation:

simple math really

6 1/2 times 0.6 = 3.9

but if you are just doing 6 and not 6 1/2 then the answer is 3.6

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