During a time of national crisis, households in America follow internet news mor

Question

During a time of national crisis, households in America follow internet news more closely than they normally do. Assume that internet viewing times are normally distributed. A group analyzing the internet viewing habits during a recent crisis sampled 16 households and produced a mean viewing time for websites of 30.2 hours per week with a standard deviation of 16.4. Use this information to answer the following questions, 1-8. 1. If the research question is, "Do the data support the idea that the mean internet viewing time exceeds the mean of 21.18 hours per week that is typically observed in American households?" then the alternative hypothesis is a statement about the mean exceeding what value? 2. State the numerical value of the test statistic that would be used to test the hypothesis that the mean is equal to 21.18 hours per week. State your answer with one digit past the decimal. 3. What is the name of the distribution that represents the set of possible test statistic values if in fact the mean internet viewing time per week during a time of national crisis is 21.18 hours per week? 4. The null hypothesis in this situation would be rejected at the 1% level if the test statistic is more than what value? 5. Assume that the value of the test statistic in this situation was 2.4. The p-value in this case is between what two values in this situation? 6. Assume the p-value in this hypothesis test is 0.07. Would the null hypothesis be rejected at the 1% significance level in this case? Answer with a YES or NO 7. Assume the p-value in this hypothesis test is 0.008. Do the data indicate that the mean internet viewing time per week during a national crisis is more than 21.18 hours at the 1% significance level stated above? Answer with a YES or NO 8. Estimate the mean viewing time for websites with a 95% confidence interval based on these data English (US)

Explanation / Answer

Solution1:

HO: mean=21.18

H1:mean 21.18

mean exceeding population mean which is 21.18

Soluton2:

t=sample mean-population mean/sample sd/.sqrt(sample size)  

=30.2-21.18/16.4/sqrt(16)

=2.2

test statistic2.2

Solution3:

t-distribution

Solution4:

n-1=16-1=15

alpha=0.05

alpha2=0.05/2=0.025

t crit =2.131

if test statistic is more than 2.131

Solution6:

p=0.07

alpha=1/100=0.01

p>level of signiifcance

p>alpha

Fail to reject null hypothesis

Accept Null hypothesis

Null hypothesis cannot be rejected.

ANSWER:NO

Solution7:

p=0.008

alpha=1%=1/100=0.01

p<alpha

Reject Null hypothesis

Accept alternativehypothesis.

ANSWER : YES

THE DATA INDICATE THAT THE MEAN INTERNET VIEWING TIME PER WEEK IS MORE THAN 21.18

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