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 130 students who averaged 38.8 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 between and texts. c. If many groups of 130 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

Answer:

Given,

standard deviation = 21.5

xbar = 38.8

n = 130

a)

Here the sampling distribution follows a T distribution

b)

Now the standard error of mean

SE = s/sqrt(n)

substitute values

= 21.5/sqrt(130)

= 1.89

Here degree of freedom = n - 1

= 130 - 1

= 129

Now foe 98% confidence interval t (0.02,129) = 2.356

Interval = xbar +/- t*SE

substitute values

= 38.8 +/- 2.356*1.89

= 38.8 +/- 4.453

= (38.8 - 4.453 , 38.8 + 4.453)

= (34.347 , 43.253)

c)

Here, 98% of these confidence intervals will contain the true population mean number of texts per day and about 2% will not contain the true population mean number of texts per day.