Suppose 100 randomly selected members of a social media site

Suppose 100 randomly selected members of a social media site were asked how much time they typically spend on the site during the week. The sample mean x was found to be 4.2 hours. Assume that time spent on the social site is approximately Normal, there are at least 5000 members and the population standard deviation is known to be =2.5 . Compute a 98% confidence interval for the average time all members of this social network spend on the site.

(a) You will use the confidence interval formula x ± z */n . Why? n

(b) Replace the variables in your formula in part (a) with the numbers for this particular confidence interval.

(c) Your answer will be: I am 98% confident that the true average number of hours a member of this social network spends on the site is between ___________and __________ hours.
(Fill in the blank with the values for the confidence interval going from small value to larger value.)

please write out steps, thank you!

Solution

a)

Because n > 30, and n < 0.05N.

b)

Note that              
              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.01          
X = sample mean =    4.2          
z(alpha/2) = critical z for the confidence interval =    2.326347874          
s = sample standard deviation =    2.5          
n = sample size =    100          
              
Thus,              
              
Lower bound =    3.618413031          
Upper bound =    4.781586969          
              
Thus, the confidence interval is              
              
(   3.618413031   ,   4.781586969   ) [ANSWER]

*******************

c)

Your answer will be: I am 98% confident that the true average number of hours a member of this social network spends on the site is between 3.618413031 and 4.781586969 hours.