10 Chi-Square Tests and the F-Distribution

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Section 10.2 143

Chi-Square Tests and

C H A P T E R

the F-Distribution

Section 10.2

4

Example 3 (pg. 526) Chi-Square Independence Test

Test the hypotheses: H o

:The number of days spent exercising per week is

10 independent of gender vs. H a

: The number of days spent exercising per week depends on gender. The first step is to enter the data in the table into Matrix A .

On the TI-83, press MATRX , highlight EDIT and press ENTER . On the TI-83

Plus, press 2 nd

[MATRX] . (MATRX is found above the x

-1

key.) Highlight

EDIT and press ENTER .

On the top row of the display, enter the size of the matrix. The matrix has 2 rows and 4 columns, so press 2 , press the right arrow key, and press 4 . Press

ENTER . Enter the first value, 40, and press ENTER . Enter the second value, 53, and press ENTER . Continue this process and fill the matrix.

Press 2 nd

[Quit] . To perform the test of independence, press STAT , highlight

TESTS , and select C:

χ 2

-Test and press ENTER .

144 Chapter 10 Chi-Square Tests and the F-Distribution

For Observed , [A] should be selected. If [A] is not already selected, press

MATRX , highlight NAMES , select 1:[A] and press ENTER . For, Expected , [B] should be selected. Move the cursor to the next line and select Calculate and press ENTER .

The output displays the test statistic and the P-value. Since the P-value is greater than

α

, the correct decision is to Fail to Reject the null hypothesis. This means that the number of days per week spent exercising is independent of gender.

Or, you could highlight Draw and press ENTER .

This output displays the

χ 2

–curve with the area associated with the P-value shaded in. The test statistic and the P-value are also displayed.

3

Section 10.2 145

4

Exercise 13 (pg. 529) Chi-Square Test for Independence

Test the hypotheses: H o

:The Result (improvement or no change) is independent of Treatment (drug or placebo) vs. H a

: The Result depends on the Treatment.

The first step is to enter the data in the table into Matrix A . Press MATRX , highlight EDIT and press ENTER . On the top row of the display, enter the size of the matrix. The matrix has 2 rows and 2 columns, so press 2 , press the right arrow key, and press 2 . Press ENTER . Enter the first value, 39, and press

ENTER . Enter the second value, 25, and press ENTER . Continue this process and fill the matrix.

Press 2 nd

[Quit] . Press STAT , highlight TESTS , and select C:

χ 2

-Test and press

ENTER . For Observed , [A] should be selected. If [A] is not already selected, press MATRX , highlight NAMES , select 1:[A] and press ENTER . For,

Expected , [B] should be selected. Move the cursor to the next line and select

Calculate and press ENTER .

The output displays the test statistic and the P-value. Since the P-value is less than

α

, the correct decision is to Reject the null hypothesis. The correct recommendation is to use the drug as part of the treatment.

3

146 Chapter 10 Chi-Square Tests and the F-Distribution

Section 10.3

4

Example 3 (pg. 537) Performing a Two-Sample F-Test

Test the hypotheses: H o

:

σ

1

2 ≤ σ

2

2

vs. H a

:

σ

1

2 > σ

2

2

. The sample statistics are: s

1

2 =

144, n

1

=

10, s

2

2 =

100 and n

2

=

21. Press STAT , highlight TESTS and select D:2-SampFTest . For Inpt , select Stats and press ENTER . On the next line, enter s

1

, the standard deviation under the old system. The standard deviation is the square root of the variance, so press 2 nd

[ ] and enter 144.

Enter n

1

on the next line. Next, enter the standard deviation under the new system by pressing 2 nd

[ ] 100. Enter n

2

on the next line. Highlight

> σ

2

for the alternative hypothesis and press ENTER . Select Calculate and press

ENTER .

The output displays the alternative hypothesis, the test statistic, the P-value and the sample statistics. Since the P-value is greater than

α

, the correct decision is to Fail to Reject the null hypothesis. There is not enough evidence to conclude that the new system decreased the variance in waiting times.

You could also select Draw and press ENTER .

Section 10.3 147

The output displays the F-distribution with the area associated with the P-value shaded in.

3

148 Chapter 10 Chi-Square Tests and the F-Distribution

4

Exercise 19 (pg. 540) Comparing Two Variances

Test the hypotheses: are: s

1

= 0.7, n

1

=

H o

:

σ

1

2 ≤ σ

2

2

25, s

2

= 0.5 and

H a

:

σ

1

2 > σ 2

vs.

2

. The sample statistics n

2

=

21. Press STAT , highlight TESTS and select D:2-SampFTest . For Inpt , select Stats and press ENTER . On the next line, enter s

1

, the standard deviation under the old procedure. Enter n

1

on the next line. Next, enter the standard deviation and sample size for the new procedure and press ENTER . Highlight

> σ

2

for the alternative hypothesis and press ENTER . Select Calculate and press ENTER .

The output displays the alternative hypothesis, the test statistic, the P-value and the sample statistics. Since the P-value is less than

α

, the correct decision is to

Reject the null hypothesis. The data supports the hospital's claim that the standard deviation of the waiting times has decreased.

3

Section 10.4 149

Section 10.4

4

Example 2 (pg. 547) Performing an ANOVA Test

Test the hypotheses: H o

:

µ

1

= µ

2

= µ

3

vs. H a

: at least one mean is different from the others. Enter the data into L1, L2 , and L3 . Press STAT , highlight

TESTS and select F:ANOVA( and press ENTER , 2 nd

[L1] , 2 nd

[L2] , 2 nd

[L3].

Press ENTER and the results will be displayed on the screen.

The output displays the test statistic, F = 6.13, and the P-value, p = .00638. Since the P-value is less than

α

, the correct decision is to Reject the null hypothesis.

150 Chapter 10 Chi-Square Tests and the F-Distribution

This indicates that there is a difference in the mean flight times of the three airlines.

The output also displays all the information that is needed to set up an ANOVA table like the table on pg. 544 of your textbook.

Between

Squares

1806.5

Degrees of

Freedom

2

Mean Squares F

903.2 903.2 / 147.3

Within 3978.2 27 147.3

Note: The TI-83 labels variation "Between" samples as variation due to the

"Factor". Also, variation "Within" samples is labeled as "Error".

The final item in the output is the pooled standard deviation, Sxp = 12.14.

3

Section 10.4 151

4

Exercise 5 (pg. 549) Performing an ANOVA Test

Test the hypotheses: H o

:

µ

1

= µ

2

= µ

3

vs. H a

: at least one mean is different from the others. Enter the data into L1, L2 , and L3 . Press STAT , highlight

TESTS and select F:ANOVA( and press ENTER , 2 nd

[L1] , 2 nd

[L2] , 2 nd

[L3]. Press ENTER and the results will be displayed on the screen.

The output displays the test statistic, F = 0.77, and the P-value, p = 0.4752. Since the P-value is greater than

α

, the correct decision is to Fail to Reject the null hypothesis. The data does not support the claim that at least one mean cost per month is different.

3

152 Chapter 10 Chi-Square Tests and the F-Distribution

4

Exercise 14 (pg. 552) Performing an ANOVA Test

Test the hypotheses: H o

:

µ

1

= µ

2

= µ

3

= µ

4

vs. H a

: at least one mean is different from the others. Enter the data into L1, L2 , L3 and L4 . Press STAT , highlight TESTS and select F:ANOVA( and press ENTER , 2 nd

[L1] , 2 nd

[L2] , 2 nd

[L3] , 2 nd

[L4]. Press ENTER and the results will be displayed on the screen.

The output displays the test statistic, F = 8.46, and the P-value, p = .0002196.

Since the P-value is less than

α

, the correct decision is to Reject the null hypothesis. The data supports the claim that the mean energy consumption for at least one region is different.

3

Technology 153

4

Technology (pg. 563) Teacher Salaries

Exercise 2: One method of testing to see if a set of data is approximately normal is to use a Normal Probability Plot. To test the data in L1 , press 2 nd

[STAT

PLOT] , select Plot 1 and turn ON Plot1. For Type , select the last icon on the second line. For Data List, press 2 nd

[L1]. For Data Axis , select X . For Mark , select the first icon. Press ZOOM and 9 to display the normal plot.

Data

Data that is approximately normally distributed will have a plot that looks fairly linear. Although this graph is not perfectly straight, it is fairly linear and it is therefore reasonable to conclude that the data is approximately normal. Repeat this process to check the other datasets for normality.

Exercise 3: Test the hypotheses: H o

σ

1

2 = σ

2

2

vs.

H a

:

σ

1

2 ≠ σ

2

2

for each pair of samples. This test requires values for s

1

and s

2

. To obtain these values, enter the data into L1, L2 and L3 . To find the standard deviation for L1 , press

STAT , highlight CALC and press L1 ENTER . To find the standard deviation for L2 , press STAT , highlight CALC , press [L2] ENTER . Repeat this process to get the standard deviation of L3 .

For each comparison, press STAT , highlight TESTS and select D:2-SampFTest .

For Inpt , select Stats and press ENTER . To compare

σ 2 and

1

σ

2

2

, enter s

1

, the sample standard deviation. Enter n

1

, the corresponding sample size, on the next line. Next, enter s

2 and n

2

. Highlight

≠ σ

2

2

for the alternative hypothesis and press ENTER . Select Calculate and press ENTER . Repeat this process to compare

σ

1

2 and

σ

3

2

. Finally compare

Exercise 4: Test the hypotheses: H o

:

µ

1

σ

2

2 and

σ

= µ

2

= µ

3

2

3

.

vs. H a

: at least one mean is different from the others. Press STAT , highlight TESTS and select

F:ANOVA( and press ENTER , 2 nd

[L1] , 2 nd

[L2] , 2 nd

[L3] . Press

ENTER and the results will be displayed on the screen.

Exercise 5: Repeat Exercises 1 - 4 using this second set of data.

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