Question

NBC claims that viewers spend a daily average of 93.1 minutes watching their content. An advertiser wants to verify this and conducts a poll of 29 random viewers who claim to watch NBC. The poll showed that this group spends a daily average of 85.8 minutes watching NBC with a standard deviation of 12.5 minutes. Use a 0.005 significance level to test the claim that the daily average amount of time NBC viewers watch NBC is less than 93.1 minutes, the average amount claimed by NBC..

The test statistic is:    (to 3 decimals)

The Critical Value is:    (to 3 decimals)

Answer 1

Solution :

This is the left tailed test,  

The null and alternative hypothesis is ,

H₀ :   = 93.1

H_a : < 93.1

Test statistic = t

= ( - ) / s / n

= (85.8 - 93.1) / 12.5 / 29

Test statistic = t = -3.145

= 0.005  

degrees of freedom = n - 1 = 29 - 1 = 28

= P(t < -2.763 ) = 0.005  

critical value = t = -2.763

Fail to reject null hypothesis

test statistic < critical value