7. The General Society Survey asked a sample of 1200 people how much time they spent watching TV each day. The mean number of hours was 3.0 with a standard deviation of 2.87. A sociologist claims that people watch a mean of 4 hours of TV per day. Do the data provide sufficient evidence to disprove the claim? Use α = .05 to test the hypothesis. a. What are your null and alternative hypotheses? b. What test is appropriate here? Why? c. What is your test statistic? d. What is your critical value? e. What is your final decision: do you reject the null or fail to reject the null?

a) and b)  Look step by step explanation

c) z(s) = - 12,07

d) z(c) = - 1,64

e) Final decision: Reject H₀

Step-by-step explanation:

We assume Normal Distribution

Data:

Sample population  n   = 1200

Sample mean         μ  =    3

Sample Standard deviation   2,87

Claim mean      μ₀  =  4

α = 0,05  then from z-table we find  z(c) = 1,64   ( critical value )

We need to develop a one tail-test to the left

Test Hypothesis

The General Society developed a survey ( in all cases that is an indication of a sample)

Null hypothesis                    H₀                      μ   =  μ₀

Alternative hypothesis        Hₐ                       μ  <  μ₀

To calculate the  z(s)

z(s) =  ( μ  -  μ₀ )/ 2,87/√n

z(s) =  ( 3 - 4 )/ 2,87/√1200

z(s) =   -1 * 34,64 / 2,87

z(s) = - 12,07

To compare  z(s) and z(c)

z(s)  <  z(c)           - 12,07 < - 1,64

z(s) is in the rejection region (quite far away) we reject H₀

Data provide enough evidence to disprove the claim

c

step-by-step explanation:

it would be see because your formula is f(x)=6x, so 6 times 1 is 6 and you go on and the best answer is c.