Question

Carry out (show all steps) the test to answer each of these questions. You may use either the classical method or the P -value method, but you must state the P -value for each problem.

3. A random sample of 15 households in a small city gives the following number of hours of television watched in one week. A marketing firm wishes to determine whether the mean amount of television watching per week is more than 8 hours at the α = 0.10 level of significance.

# hours of television watched 7.5, 6.3, 7.2, 5.0, 7.9,10.7, 8.9, 8.8, 10.3, 8.8, 9.5, 9.5, 6.1, 9.4, 8.4

Answer 1

Solution:

x	x2
7.5	56.25
6.3	39.69
7.2	51.84
5	25
7.9	62.41
10.7	114.49
8.9	79.21
8.8	77.44
10.3	106.09
8.8	77.44
9.5	90.25
9.5	90.25
6.1	37.21
9.4	88.36
8.4	70.56
∑x=124.3	∑x2=1066.49

Mean ˉx=∑xn

=7.5+6.3+7.2+5+7.9+10.7+8.9+8.8+10.3+8.8+9.5+9.5+6.1+9.4+8.4/15

=124.315

=8.2867
Mean = 8.29

Sample Standard deviation S=√∑x2-(∑x)2nn-1

=√1066.49-(124.3)21514

=√1066.49-1030.0327/14

=√36.4573/14

=√2.6041

=1.6137

Sample Standard deviation =1.61

This is the two tailed test .

The null and alternative hypothesis is ,

H₀ :    = 8

H_a : > 8

Test statistic = t

= ( - ) / S / n

= (8.29-8) / 1.61 / 15

= 0.698

Test statistic = t =  0.698

P-value =0.2484

= 0.10

P-value >

0.2484 > 0.10

Fail to reject the null hypothesis .

There is not sufficient evidence to claim that the population mean μ is greater than 8, at the 0.05 significance level.