# Suppose an ad agency claims th

Suppose an ad agency claims that likes per hour of their ad isnot equal to 7 likes, on average. Several of their employees do notbelieve them, so a manager decides to do a hypothesis test, at a 5%significance level, to persuade them. She collects data for 19hours 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 theresults.

Reject H0. At the 5% significance level, the test results arenot statistically significant and at best, provide weak evidenceagainst the null hypothesis.

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

Do not reject H0. At the 5% significance level, the test resultsare not statistically significant and at best, provide weakevidence against the null hypothesis.

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

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 providesufficient evidence to conclude that the mean likes per hour oftheir ad is not equal to 7 likes.

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