The average duration of labor from the first contraction to the birth of the baby in...

It is supposed that a sample of n=29 women exercise daily and have an average duration of labor of 17.8 hours, and sample variance of 77.4 hours. Thus, 1 =17.8, and s2 = 77.4. Also given, the average duration of labor is 16 hours, that is, j = 16. The estimated standard error (sv) is derived as follows: s 177.4 V29 = 1.634 The t statistic is obtained as follows: tru 17.8-16 V77.4 29 17.8-16 1.634 = 1.1016 The estimated standard error is sy = 1.634).

The t statistic is 1.1016).

When the sample size is taken larger, that is, n= 73. The sample average and the sample standard deviation is 1 =17.8, and s? = 77.4, respectively. The estimated standard error (sw) is derived as follows: su In 773 = 1.030 The t statistic is obtained as follows: - t= 2 S 17.8-16 V77.4 V73 17.8-16 1.030 = 1.7476 The new estimated standard error is sy = 1.030. The new t statistic is 1.7476 The t statistic for larger n= 73 is t = 1.7476.

The t statistic for smaller n= 29 is t = 1.1016. Thus, as n becomes larger, the t statistics become larger and as n becomes smaller, the t statistics become smaller. Hence, the t statistic becomes larger as n becomes larger.

The probability of getting the t statistic or something more extreme for the sample size of n=29, that is, the p-value is obtained as follows: As the null hypothesis is that the average duration of labor is 16 hours, that is, j = 16. The alternative hypothesis against the null hypothesis is that the average duration of labor is not equal to 16 hours, that is, u z 16.

(two-tailed test). The degrees of freedom are df =(n-1)=(29-1) = 28. The t statistic as derived above is t = 1.1016. The p-value is calculated as follows: p-value = 2P(t>1.1016) = 2[1- P(t51.1016)] = 2/1 -0.8643] = 2x0.1357 2x0.14 = 0.28 (as n=29 > 25, using standard normal table) Thus, p = 0.28

The probability of getting the t statistic or something more extreme for the sample size of n= 73, that is, the p-value is obtained as follows: The degrees of freedom are df =(n-1)=(73 - 1) = 72. The t statistic as derived above is t = 1.7476.

The p-value is calculated as follows: p-value = 2P(t >1.7476) = 2[1- P(t 51.7476)] = 2 [1–0.9599] = 2x0.0401 2x0.04 = 0.08 (as n= 73 > 25, using standard normal table) Thus, p=0.08. The results obtained are as follows: For n= 29, the estimated standard error is sx = 1.634). The t statistic is 1.1016. For n= 73, the new estimated standard error is sy = 1.030. The new t statistic is 1.7476.

The t statistic becomes larger as n becomes larger. For n = 29, p = 0.28. For n=73, p=0.08) The t distribution is decreasing with a smaller n.