Question

One significant consequence of the COVID-19 situation is that many meetings that formerly were held face-to-face are now held virtually using something like Zoom or Skype. Suppose the first population is all face-to-face meetings held in June 2020, the second population is all Zoom meetings held in June 2020, and the parameter of interest is 41-42 the difference in the mean length of all face-to-face meetings and the mean length of all Zoom meetings. The meeting lengths are measured in minutes. A simple random sample of 72 face-to-face meetings held in June 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 June 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

Sample Sample mean : = 48 Given 72 n = S1 Given Given 14.3 Sample size : Sample standard deviation: Sample 11 Sample mean : X2 53 Given 61 n2 = Given Sample size Sample standard deviation : 12.8 Given Point estimate : 11 - 12 = -5 2 2 S2 Standard error + 2.350753 n n2 Level of significance : a= 0.05 Given 95% confidence level + 2 2 S1 S2 nu n2 deg rees of freedom= (salma (s321m2 :. Critical values : tc/2 = + 1.978 131 + 7-1 12-1 S2 Margin of error: ta/2 S1 n + 11 4.65 112 Lower bound: (51 - 12)-tal2, 2 S1 2 S2 + -9.650684 n n2 2 $1 2 S2 Upper bound: (ī - 2 ) +ta/24 + = -0.349316 n n2 The confidence interval is (-9.651,-0.349)