Question

one hundred eight americans were surveyed determine. the number of hours they spend watching tv each month. it was revealed that they watched an average of 151hours each month with a standard deviation of 32 hours. Assume that the underlying population destitution is normal.

b. find critical value of 99% population mean time spent waiting

c. which distribution should you use for this problem?

d.calculate error bound

e. sketch graph

f. what are the lower and upper bounds

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 151  

Population standard deviation = s = 32

Sample size = n = 108

b) At 99% confidence level

= 1-0.99% =1-0.99 =0.01

/2 =0.01/ 2= 0.005

t/2,df = t0.005,107 = 2.62

critical value = t/2,df = 2.62

c)singal sampla t value confidence interval useing

d) Margin of error = E = t/2,df * (s /n)

= 2.62 * (32 / 108)

Margin of error = E = 8.08

The 99% confidence interval estimate of the population mean is,

- E < <  + E

151 - 8.08 < < 151 + 8.08

142.92 < < 159.08

(142.92 ,159.08)

e)

f)

lower bound = 142.92

upper bound = 159.08