Use Stat Tests T Interval

Question

A sample of  150 college students at a large university reports getting an average of 6.55 hours of sleep last night with a standard deviation of 1.82 hours.

       1. Use a t distribution to construct a 98% confidence interval for the average amount of sleep students at this university got last night. Use two decimal places in your margin of error. Use Stat Tests T Interval.

           

                       A) 4.73 to 8.97 hours

                       B) 6.14 to 7.96 hours

                       C) 5.32 to 8.14 hours

                       D) 6.20 to 6.90 hours

Answer 1

The interval is: x_av ± t_c(s/√n);

where x_av = 6.55 hours, s = 1.82 hours, n = 150

t_c = 2.3545 for 98% confidence and for degree of freedom 150 - 1 = 149

So, interval is: 6.55 ± 2.3545(1.82/√150) = 6.55 ± 0.35

Left hend end point of interval is 6.55 - 0.35 = 6.20 hours;

Right end point of interval is 6.55 + 0.35 = 6.90 hours

98% confidence interval for μ: (6.20, 6.90):