Question

A simple random sample of 72 face-to-face meetings held in March 2020 was selected, and the mean length of this sample of 72 meetings was 48 minutes with a standard deviation of 14.3 minutes. An independent simple random sample of 61 Zoom meetings held in March 2020 was selected, and the mean length of this sample of 61 meetings was 53 minutes with a standard deviation of 12.8 minutes. If appropriate, use this information to calculate and interpret a 95% confidence interval for the difference in the mean length of all face-to-face meetings and the mean length of all Zoom meetings.

Answer 1

Answer:

The 95% confidence interval for the difference in the mean length of all face-to-face meetings and the mean length of all Zoom meetings is (-9.70, 0.31).

Step-by-step explanation:

The (1 - α)% confidence interval for the difference between two means, in case of unknown population standard deviation is:

$CI=(\bar x_{1}-\bar x_{2})\pm t_{\alpha/2, (n_{1}+n_{2}-2)}\times S_{p}\times \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}$

The information provided is:

$n_{1}=72\\\bar x_{1}=48\\s_{1}=14.3\\n_{2}=61\\\bar x_{2}=53\\s_{2}=12.8$

Compute the value of pooled standard deviation as follows:

$S_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}}=\sqrt{\frac{(72-1)14.3^{2}+(61-1)12.8^{2}}{72+61-2}}=13.63$

Compute the critical value of t as follows:

$t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.05/2, (72+61-2)}=t_{0.025, 131}=1.978$

*Use a t-table.

Compute the 95% confidence interval for the difference between two means as follows:

$CI=(\bar x_{1}-\bar x_{2})\pm t_{\alpha/2, (n_{1}+n_{2}-2)}\times S_{p}\times \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}$

     $=(48-53)\pm 1.978\times 13.63\times\sqrt{\frac{1}{72}+\frac{1}{61}}\\=-5\pm 4.6916\\=(-9.6916, -0.3084)\\\approx (-9.70, -0.31)$

Thus, the 95% confidence interval for the difference in the mean length of all face-to-face meetings and the mean length of all Zoom meetings is (-9.70, 0.31).