Question

The following is a random sample of the annual salaries of high school counselors in the United States. Assuming that the distribution of salaries is approximately normal, construct a 98% confidence interval for the mean salary of high school counselors across the United States. Round to the nearest dollar. $45,860,$38,860,$64,820,$63,480,$36,710,$50,410,$33,080

Answer 1

Solution :

      x                   $(x-\overline x)$                          $(x-\overline x)^2$

45860             -1742.8571                 3037551.0204

38860            -8742.8571                 76437551.0204

64820              17217.1429                296430008.1633

63480             15877.1429                252083665.3061

36710              -10892.8571              118654336.7347

50410             2807.1429                 7880051.0204

33080             -14522.8571            210913379.5918    

333220          0.0000                        965436542.8571

Sample size, n = 7

Mean = $\frac{\sum x}{n}=\frac{333220}{7}$

                   = 47602.8571

Variance = $\frac{(\sum (x- \overline x))^2}{(n-1)}=\frac{965436542.8571}{7-1}$

                                       = 160906090

Standard deviation = $\sqrt{Variance} = \sqrt{160906090}$

                                                     = 12684.876

a). df = n - 1

        = 7 - 1

       = 6

Level of significance, α = 0.02

Critical, $t_c = 3.143$

b). Sample mean, $\overline x = 47602.8571$

    Sample standard deviation, s = 12684.876

Sample size, n = 7

c). 98% confidence interval = $\overline x \pm t_c \times \frac{s}{\sqrt n}$

                                             $=47602.8571 \pm 3.143 \times \frac{12684.876}{\sqrt 7}$

                                             $=(32533.96,62671.76)$