Question

NBC claims that viewers spend a daily average of 175.1 minutes watching their content. An advertiser wants to verify this and conducts a poll of 10 random viewers who claim to watch NBC. The poll showed that this group spends a daily average of 163.3 minutes watching NBC with a standard deviation of 15.7 minutes. Use a 0.025 significance level to test the claim that the daily average amount of time NBC viewers watch NBC is at least 175.1 minutes, the average amount claimed by NBC..

The test statistic is:                             [ Select ]                       ["1.87", "does not", "-3.51", "-1.55", "3.25", "does", "-4.21", "-2.38"]      

The Critical Value is:  (to 4 decimals)                             [ Select ]                       ["-2.2621", "3.2578", "-3.2247", "2.4478", "-1.8896"]      

Based on this we:                            [ Select ]                       ["Reject the null hypothesis", "Fail to reject the null hypothesis"]      

Conclusion There [conclusion]  appear to be enough evidence to support the claim that the daily average amount of time NBC viewers watch NBC is at least 175.1 minutes, the average amount claimed by NBC..

Answer 1

Sol;

Ho:m>=175.1

Ha;mu<175.1

t=xbar-mu/s/sqrt9n)

t=(163.3-175.1)/(15.7/sqrt(10))

t=-2.376744

t=-2.38

critical t value in excel

df=n-1=10-1=9

=T.INV(0.025,9)

=-2.262157163

based on this we reject Ho

ANSWER;

The test statistic is: -2.38

The Critical Value is:-2.2622

test statistc<critical value

Based on this we:       reject the null hypothesis"]      

There appear to be enough evidence to support the claim that the daily average amount of time NBC viewers watch NBC is at least 175.1 minutes, the average amount claimed by NBC..