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Statistics 210. Problem 9.68 May 2, 2013. Problem # 9.68.

Statistics 210

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Statistics 210

Problem 9.68

May 2, 2013

Problem # 9.68

A past Study claimed that adults Americans spent an average of 18 hours/week on leisure activity. A researcher wanted to test this claim. She took 12 adults and asked them how much time per week they spent on leisure time activities, with the following resultsAssume that time spent by all adults is normally distributedUsing a 10% significance level, Can you conclude that that the average time spent on leisure activities has changed?(Hint: First calculate the sample mean and the sample standard deviation for these data using formulas learned in Sections 3.1.1 and 3.2.2)

The following are what we will call the basic formulas that are used to calculate the variance:

Where σ2 is the population variance and s2 is the sample variance.

The standard deviation is obtained by taking the positive square root of the variance.

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From the computational point of view, it is easier and more efficient to use short-cut formulasto calculate the variance and standard deviation. By using the short-cut formulas, we reduce the computation time and round-off errors.

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- ∑x = 248.1
- ∑x2= 5633.53
- n = 12
- X bar (sample mean) = 20.675 (248.1/12)
- Sample Standard Deviation = 6.76933

α√µ∑

∑

- State the null and alternative hypotheses;
- Select the appropriate distribution;
- Determine the rejection and non-rejection regions.
4. Then calculate the p-value and compare it to the given Significance Level.

- P value > Alpha = Cannot reject
- P Value < Alpha = Reject

- State the null and alternative hypotheses:
- H0 about the mean: Leisure Time = 18 hrs/week

- H1 about the mean: Leisure Time ≠ 18 hrs/week ….. Thus a Two-tailed test

- Select the appropriate distribution;
- 10% Significance Level = helps us determine the “rejection area”

- Test of statistic:

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Case I. If the following three conditions are fulfilled:

1.The population standard deviation σ is not known

2. The sample size is small (i.e., n < 30)

3. The population from which the sample is selected is normally distributed,………..

then we use the t distribution to perform a test of hypothesis about μ.

- Determine the rejection and non-rejection regions.
- 10% Significance Level, .05 area under the curve on each tail.

4. Then calculate the p-value or the test statistic to complete the hypothesis test.

- P value > Alpha = Cannot reject
- P Value < Alpha = Reject

.05

.05

α√µ∑

- Hypothesized Mean (µ)
- n = 12
- X (sample mean) = 20.675
- Sample Standard Deviation = 6.76933

α√µ∑

∑

Case I. If the following three conditions are fulfilled:

1.The population standard deviation σ is not known

2. The sample size is small (i.e., n < 30)

3. The population from which the sample is selected is normally distributed,………..

then we use the t distribution to perform a test of hypothesis about μ.

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