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 103 students who averaged 27.9 texts per day. The standard deviation was 15.9 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 103 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)

To compute the confidence interval use a Z distribution because sample size is large enough (n≥30)

b)

std dev , s =    15.9000
Sample Size ,   n =    103
Sample Mean,    x̅ =   27.9000
      
Level of Significance ,    α =    0.1          
'   '   '          
z value=   z α/2=   1.6449   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = s/√n =   15.900   / √   103   =   1.5667
margin of error, E=Z*SE =   1.6449   *   1.567   =   2.577
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    27.90   -   2.577   =   25.3231
Interval Upper Limit = x̅ + E =    27.90   -   2.577   =   30.4769
90%   confidence interval is (   25.32   < µ <   30.48   )

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.