Question

In a random sample of 80 teenagers, the average number of texts handled in one day is 50. The 96% confidence interval for the mean number of texts handled by teens daily is given as 46 to 54. a. What is the standard deviation of the sample? b. If the number of samples were doubled

Answer 1

Standard deviation of the sample = 17.421

Explanation:

We are given that in a random sample of 80 teenagers, the average number of texts handled in one day is 50.

Also, the 96% confidence interval for the mean number of texts handled by teens daily is given as 46 to 54.

So, sample mean,
`xbar` = 50 and Sample size, n = 80

Let sample standard deviation be s.

96% confidence interval for the mean number of texts,
`\mu` is given by ;

96% confidence interval for
`\mu` =
`xbar \pm 2.0537*(s)/(√(n) )`

[46 , 54] =
`50 \pm 2.0537*(s)/(√(80) )`

Since lower bound of confidence interval = 46

So, 46 =
`50 - 2.0537*(s)/(√(80) )`

50 - 46 =
`2.0537*(s)/(√(80) )`

s =
`(4*√(80) )/(2.0537)` = 17.421

Therefore, standard deviation of the sample is 17.421 .