Question

The average duration of labor from the first contraction to the birth of the baby in women over 35 who have not Use the Distributions tool to look at the t distributions for different sample sizes. To do this, choose the Degrees of previously given birth and who did not use any pharmaceuticals is 16 hours. Freedom for the first sample size on the slider, and click the radio button with the single orange line. Move the orange vertical line to the right until the number below the orange line is located on the t statistic. The probability of Suppose you have a sample of 29 women who exercise daily, and who have an average duration of labor of 17.8 getting that t statistic or one more extreme will appear in the bubble with the orange type. Now repeat the process for the other sample. 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. t Distribution Calculate the t statistic. To do this, you first need to calculate the estimated standard error. The estimated standard error is SM= .The t statistic is Degrees of Freedom 52 Now suppose you have a larger sample size n 73. Calculate the estimated standard error and thet statistic for thi sample with the same sample average and the same standard deviation as above, but with the larger sample size. . The new t statistic is The new estimated standard error is Note that the t statistic becomes as n becomes larger. - 2.0 2.0 3.0 -1.0 1.0 3.0 AA What is the probability of getting the t statistic or something more extreme for the sample size of n 29? p What is the probability of getting the t statistic or something more extreme for the sample size of n 73? p The t distribution is with a smaller n. (Hint: To best see this, click the radio button in the tool with no vertical lines. Slowly move the Degrees of Freedom slider from the smallest value to the largest value, and observe how the shape of the distribution changes.)

Answer 1

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.