Question

A researcher is interested in finding a 90% confidence interval for the mean number of times per day that college students text. The study included 142 students who averaged 36.7 texts per day. The standard deviation was 21.9 texts. Round answers to 3 decimal places where possible. a. To compute the confidence interval use a distribution. b. With 90% confidence the population mean number of texts per day is between and texts. c. If many groups of 142 randomly selected members are studied, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population number of texts per day and about percent will not contain the true population mean number of texts per day.

Answer #1

(a)

**t distribution**

(b)

SE = s/

= 21.9/

= 1.8378

= 0.10

ndf = n - 1 = 142 - 1 = 141

From Table, critical values of t = 1.6557

Confidence Interval:

36.7 (1.6557 X 1.8378)

= 36.7 3.0429

= ( 33.6571 ,39.7429)

Confidence interval:

**33.657 < <
39.743**

(c)

About **90%** percent of these confidence intervals
will contain the true population number of texts per day and about
**10%** percent will not contain the true population
mean number of texts per day

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