Question

A researcher is interested in finding a 95% confidence interval for the mean number of times per day that college students text. The study included 144 students who averaged 35.8 texts per day. The standard deviation was 23.6 texts. Round answers to 3 decimal places where possible. a. To compute the confidence interval use a distribution. b. With 95% confidence the population mean number of texts per day is between and texts. c. If many groups of 144 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

Solution :

Given that,

a) Point estimate = sample mean = = 35.8

sample standard deviation = s = 23.6

sample size = n = 144

Degrees of freedom = df = n - 1 = 144 - 1 = 143

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t0.025,143 = 1.977

Margin of error = E = t/2,df * (s /n)

= 1.977 * ( 23.6 / 144)

Margin of error = E = 3.89

The 95% confidence interval estimate of the population mean is,

  ± E

= 35.8   ± 3.89

= ( 31.91, 39.69 )

b) With 95% confidence the population mean number of texts per day is between 31.91 and 39.69 texts.

c) If many groups of 144 randomly selected members are studied, then a different confidence interval would be produced from each group. About 95 percent of these confidence intervals will contain the true population number of texts per day and about 5 percent will not contain the true population mean number of texts per day.