Question

A researcher is interested in finding a 98% confidence interval for the mean number of times per day that college students text. The study included 134 students who averaged 25.9 texts per day. The standard deviation was 21.5 texts. Round answers to 3 decimal places where possible.

a. To compute the confidence interval use a distribution.

b. With 98% confidence the population mean number of texts per day is answer ___ between and answers ___ texts.

c. If many groups of 134 randomly selected members are studied, then a different confidence interval would be produced from each group. About answer ___ percent of these confidence intervals will contain the true population number of texts per day and about answer ___ percent will not contain the true population mean number of texts per day.

Answer #1

a)

use z distribution

b)

sample mean, xbar = 25.9

sample standard deviation, σ = 21.5

sample size, n = 134

Given CI level is 98%, hence α = 1 - 0.98 = 0.02

α/2 = 0.02/2 = 0.01, Zc = Z(α/2) = 2.33

ME = zc * σ/sqrt(n)

ME = 2.33 * 21.5/sqrt(134)

ME = 4.33

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))

CI = (25.9 - 2.33 * 21.5/sqrt(134) , 25.9 + 2.33 *
21.5/sqrt(134))

CI = (21.572 , 30.228)

With 98% confidence the population mean number of texts per day is 21.572 between and 30.228 texts.

c)

If many groups of 134 randomly selected members are studied, then a
different confidence interval would be produced from each group.
About answer98 percent of these confidence intervals will contain
the true population number of texts per day and about answer 2
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