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 137 students who averaged 40.7 texts per day. The standard deviation was 13.4 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 137 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)

since, sample size is large enough, z distribution will be used

sample std dev ,    s =    13.4000
Sample Size ,   n =    137
Sample Mean,    x̅ =   40.7000

b)

Level of Significance ,    α =    0.02          
'   '   '          
z value=   z α/2=   2.3263   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = s/√n =   13.400   / √   137   =   1.1448
margin of error, E=Z*SE =   2.3263   *   1.145   =   2.663
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    40.70   -   2.663   =   38.0367
Interval Upper Limit = x̅ + E =    40.70   -   2.663   =   43.3633
98%   confidence interval is (   38.037 < µ <   43.363 )

c)

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