Statistics 210

Problem 9.68May 2, 2013

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 results

Assume that time spent by all adults is normally distributed

Using 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)

Survey Responses

1 13.62 143 24.54 24.65 22.96 37.77 14.68 14.59 21.5

10 2111 17.812 21.4

n=12 248.1Sample Mean 20.675

Problem # 9.68

Section 3.1.1 and 3.2.2Mean and Standard Deviation

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.

From the computational point of view, it is easier and more efficient to use short-cut formulas to 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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Section 3.1.1 and 3.2.2Mean and Standard Deviation

• ∑x = 248.1• ∑x2 = 5633.53• n = 12• X bar (sample mean) = 20.675 (248.1/12)

• Sample Standard Deviation = 6.76933

α√µ∑∑

General Procedures

1. State the null and alternative hypotheses;2. Select the appropriate distribution; 3. 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

General Procedures

1. 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

General Procedures2. Select the appropriate distribution;

- 10% Significance Level = helps us determine the “rejection area”- Test of statistic:

Case I. If the following three conditions are fulfilled:1. The population standard deviation σ is not known2. 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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General Procedures

3. 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

α√µ∑

.05.05

Calculate T value and reference T-table using degrees of freedom (df)

• Hypothesized Mean (µ)

• n = 12

• X (sample mean) = 20.675

• Sample Standard Deviation = 6.76933α√µ∑∑

Result

Thank You

General Procedures9.3 Hypothesis Tests About μ: σ Not Known

Case I. If the following three conditions are fulfilled:1. The population standard deviation σ is not known2. 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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Survey Responses

X x-bar

1 13.6 20.675 -7.075

2 14 20.675 -6.675

3 24.5 20.675 3.825

4 24.6 20.675 3.925

5 22.9 20.675 2.225

6 37.7 20.675 17.025

7 14.6 20.675 -6.075

8 14.5 20.675 -6.175

9 21.5 20.675 0.825

10 21 20.675 0.325

11 17.8 20.675 -2.875

12 21.4 20.675 0.725

n=12 248.1 -7.10543E-15 Sum x -xbar squared = 50.487135

Sample Mean 20.675