SOLUTION: a social scientist claims that the average adult watches less than 26 hours of television per week. he collects data on 25 individuals television viewing habits and finds that the

Algebra ->  Probability-and-statistics -> SOLUTION: a social scientist claims that the average adult watches less than 26 hours of television per week. he collects data on 25 individuals television viewing habits and finds that the       Log On


   



Question 861875: a social scientist claims that the average adult watches less than 26 hours of television per week. he collects data on 25 individuals television viewing habits and finds that the mean number of hours that the 25 people spent watching television was 22.2 hours. if the population standard deviation is known to be eight hours, can we conclude at the 1%significance level that he is right?
(a)state the appropriate null and alternative hypis be rejected or notothesis
(b)assuming that your data are normally distributed, select an appropriate statistical test and calculate the test statistic
(c)determine the critical region at the @=0.1 level of significance
(d)calculate the p-value of the test

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
a social scientist claims that the average adult watches less than 26 hours of television per week. he collects data on 25 individuals television viewing habits and finds that the mean number of hours that the 25 people spent watching television was 22.2 hours. if the population standard deviation is known to be eight hours, can we conclude at the 1%significance level that he is right?
(a)state the appropriate null and alternative hypis be rejected or notothesis
Ho: u >= 26
Ha: u < 26 (claim)
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(b)assuming that your data are normally distributed, select an appropriate statistical test and calculate the test statistic
t(22.2) = (22.2-26)/[i/sqrt(22)] = -3.8/[8/sqrt(22)] = -2.2279
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(c)determine the critical region at the @=0.1 level of significance
z = invNorm(0.10) = -1.2816
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(d)calculate the p-value of the test
p-value = P(t < -1.2816) = tcdf(100,-1.2816 with df = 21)= 0.1070
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Cheers,
Stan H.
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