Question

Suppose an ad agency claims that likes per hour of their ad is not equal to 7 likes, on average. Several of their employees do not believe them, so a manager decides to do a hypothesis test, at a 5% significance level, to persuade them. She collects data for 19 hours and works through the testing procedure:

• H0: μ=7; Ha: μ≠7
• x¯=6.2
• σ=0.9
• α=0.05 (significance level)
• The test statistic is

  z0=x¯−μ0σn√=6.2−70.919√=−3.87

• The critical values are −z0.025=−1.96 and z0.025=1.96.

Conclude whether to reject or not reject H0, and interpret the results.

Select the correct answer below:

Reject H0. At the 5% significance level, the test results are not statistically significant and at best, provide weak evidence against the null hypothesis.

Reject H0. At the 5% significance level, the data provide sufficient evidence to conclude that the mean likes per hour of their ad is not equal to 7 likes.

Do not reject H0. At the 5% significance level, the test results are not statistically significant and at best, provide weak evidence against the null hypothesis.

Do not reject H0. At the 5% significance level, the data provide sufficient evidence to conclude that the mean likes per hour of their ad is not equal to 7 likes.

Answer 1

Solution :

= 7

= 6.2

= 0.9

n = 19

This is the two tailed test .

The null and alternative hypothesis is

H0 :   = 7

Ha : 7

Test statistic = z

= ( - ) / / n

= ( 6.2 - 7) /0.9 / 19

= -3.87

P (Z < -3.87) =0.0002

P-value = 0.0002

= 0.05  

Critical value = 1.96

0.0002< 0.05

Reject H0. At the 5% significance level, the data provide sufficient evidence to conclude that the mean likes per hour of their ad is not equal to 7 likes.