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# Testing Hypotheses and t Distribution Statistics

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3. An earlier study claims that U.S. adults spend an average of 114 minutes with their families per day. A recently taken sample of 25 adults showed that they spend an average of 109 minutes per day with their families. The sample standard deviation is 11 minutes. Assume that the time spent by adults with their families has an approximate normal distribution. We wish to test whether the mean time spent currently by all adults with their families is less than 114 minutes a day.

a) Construct a 95% confidence interval for the mean tine spent by all adults with their families.

b) Does the sample information support that the mean time spent currently by all adults with their families is less than 114 minutes a day? Explain your conclusion in words.

2. A wine manufacturer sells Cabernet Sauvignon with a label that asserts an alcohol content of 11%. Sixteen bottles of this cabernet are randomly selected and analyzed for alcohol content. The resulting observations are:
10.8, 9.6, 9.5, 11.4, 9.8, 9.1, 10.4, 10.7, 10.2, 9.8, 10.4, 11.1, 10.3, 9.8, 9.0, 9.8
The manufacturer claims that the mean alcohol content of its cabernet sauvignon is 11%. At the 98% level of confidence, can we conclude that the claim is true?

3. A shop manual gives 6.5 hours as the average time required to perform a 30,000 mile major maintenance service on a Porsche 911. Last month a mechanic performed 11 such services, and his required times were as follows.

6.3 6.6 6.7 5.9 6.3 6.0 6.5 6.1 6.2 6.4 6.3

At the 95% level of confidence, can we conclude that the mechanic can perform this service in less time than specified by the service manual?

https://brainmass.com/statistics/hypothesis-testing/testing-hypotheses-tdistribution-statistics-505642

#### Solution Preview

1a) The critical value for 95% confidence interval is 2.06 (TINV(0.05,24), TINV is a function in EXCEL).
Margin of error=critical value*standard deviation/sqrt(n)=2.06*11/sqrt(25)=4.5
The upper limit: 109+4.5=113.5
The lower limit: 109-4.5=104.5
Therefore, the 95% confidence interval for the mean time is [104.5min, 113.5min].

1b) Since all the values in this 95% confidence interval is less than 114, we could conclude that the mean time spent currently by all ...

#### Solution Summary

The solution discusses testing hypotheses and t-distribution statistics.

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