A market researcher for a consumer electronics company wants to determine if the
ID: 3041593 • Letter: A
Question
A market researcher for a consumer electronics company wants to determine if the residents of a particular city are spending more time watching TV than the average for this geographic area. The average for this geographic area is 13 hours per week. A random sample of 16 respondents of the city is selected, and each respondent is instructed to keep a detailed record of all television viewing in a particular week. For this sample the viewing time per week has a mean of 15.3 hours and a sample standard deviation sn = 3.8 hours. Assume that the amount of time of television viewing per week is normally distributed. Can the researcher claim that the residents of this particular city are spending significantly (at 5% level) more time watching TV than the average for this geographic area? Explain your conclusion.
Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: < 13
Alternative hypothesis: > 13
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.
Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).
SE = s / sqrt(n)
S.E = 0.95
z = (x - ) / SE
z= 2.42
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
The observed sample mean produced a z statistic test statistic of 2.42.
Thus the P-value in this analysis is 0.0078.
Interpret results. Since the P-value (0.0078) is less than the significance level (0.01), we have to reject the null hypothesis.
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