Question

3) A manufacturer would like to determine if the variance in a product dimension exceeds 10. a) State the Hypothesis to show the variance is greater than 10. b) Choose a level of a Use a= 0.05 for this problem. c) To test the hypothesis, the manufacturer takes a sample of 18 parts and measure that product dimension. The data appear in the Dimension worksheet of the HW3 data workbook on Moodle. Collect data and calculate necessary statistics to test the hypothesis. d) Sketch the sampling distribution. Include the critical value and test statistic. e) Draw a conclusion and report that in the problem context. f) What is the p-value for the hypothesis test?

Part	Data
1	48.3
2	48.5
3	54.0
4	49.3
5	48.8
6	56.6
7	45.8
8	55.2
9	45.0
10	52.0
11	46.4
12	51.2
13	50.4
14	47.6
15	45.1
16	44.8
17	44.9
18	56.3

Answer 1

Solution:

Given: A manufacturer would like to determine if the variance in a product dimension exceeds 10.

Sample size = n = 18

Part a) State the Hypothesis to show the variance is greater than 10.

$H_{0}: \sigma^{2}=10$

Vs

$H_{1}: \sigma^{2}>10$

Part b) Choose a level of a Use a= 0.05 for this problem

Find critical value for a significance level = 0.05

df = n - 1 = 18 - 1 = 17

x 00 x950 x975 x900 X990 df 050 995 3.841 5.991 7,815 9.488 11.070 0,001 1 0.000 0.000 0.004 0.016 2.706 0.020 2 0.010 0.051 0.103 0.211 4.605 0.072 0.207 6.251 2 0.115 0.216 0.352 0.584 4 0.297 0.484 0.711 1.064 7.779 0.412 0.554 0.831 1.145 1.610 9.236 0.872 1.237 2.204 10.645 12.592 14 067 15.507 16.919 18.307 19.675 21.026 22.362 23 685 24996 26 296 6 0.676 1.635 0.989 1.239 1.690 2.167 2.833 12.017 1.344 1.646 2.180 2.733 3.490 13.362 1.735 2.088 2.700 3.325 4.168 14.684 2.156 15,987 10 2.558 3.247 3.940 4.865 11 2.603 3.053 3.816 4.575 5.578 17.275 3.074 12 3.571 4.404 5.226 6.304 18.549 13 3.565 4.107 5.009 5.892 7.042 19.812 14 4.075 4.660 5.629 6.571 7.790 21.064 15 4.601 5.229 6.262 7.261 8.547 22.307 7.962 5.812 16 5.142 6.908 9.312 23.542 8.672 10.085 24.769 17 27.587 7.564

Chi-square critical value = 27.587

$\chi ^{2}_{crit}=27.587$

Part c) Find sample variance and Chi-square test statistic for variance.

$\chi ^{2}=\frac{(n-1) \times s^{2}}{\sigma^{2}}$

where

$s^{2}= \frac{\sum x^{2}-(\sum x)^{2}/n}{n-1}$

Thus we need to make following table:

Part	Data x	x 2
1	48.3	2332.89
2	48.5	2352.25
3	54.0	2916
4	49.3	2430.49
5	48.8	2381.44
6	56.6	3203.56
7	45.8	2097.64
8	55.2	3047.04
9	45.0	2025
10	52.0	2704
11	46.4	2152.96
12	51.2	2621.44
13	50.4	2540.16
14	47.6	2265.76
15	45.1	2034.01
16	44.8	2007.04
17	44.9	2016.01
18	56.3	3169.69
	$\sum x =890.2$	$\sum x^{2} =44297.38$

Thus we get:

$s^{2}= \frac{\sum x^{2}-(\sum x)^{2}/n}{n-1}$

$s^{2}= \frac{44297.38 -(890.2)^{2}/18}{18-1}$.

$s^{2}= \frac{44297.38 -44025.33556 }{17}$

$s^{2}= \frac{272.04444 }{17}$

$s^{2}=16.00261$

Thus we get:

$\chi ^{2}=\frac{(n-1) \times s^{2}}{\sigma^{2}}$

$\chi ^{2}=\frac{(18-1) \times 16.00261}{10}$

$\chi ^{2}=\frac{17 \times 16.00261}{10}$

$\chi ^{2}=27.204$

Part  d) Sketch the sampling distribution. Include the critical value and test statistic.

Rejection Region 0.05 227.204 Xerit 2 27.587

Part e) Draw a conclusion and report that in the problem context.

Decision rule: Reject null hypothesis H0, if Chi-square test statistic value > Chi-square critical value = 27.587, otherwise we fail to reject H0.

Since Chi-square test statistic value =$\chi ^{2}=27.204$ < Chi-square critical value = 27.587,we fail to reject H0.

Thus there is not sufficient evidence to conclude that: the variance in a product dimension exceeds 10.

Part f) What is the p-value for the hypothesis test?

To get exact p-value, use Excel command:

=CHISQ.DIST.RT( x² , df )

=CHISQ.DIST.RT(27.204 , 17 )

=0.0551

p-value = 0.0551

To get interval of p-value using table, look in Chi-square table for df = 17 row and find interval in which $\chi ^{2}=27.204$ fall, then find corresponding right tail area interval , which would be range of p-value.

x995 x373 x950 x300 xoso df .990 100 .050 .025 3841 5991 7815 9 488 11.070 12 592 14 067 15507 16919 18 307 19 675 21 026 22.362 23.85 24.996 26.396 2.06 4.605 6251 7.179 9.236 10. 45 12.017 13862 14.684 15987 17275 18549 19812 21 064 22 307 23 42 1 0.000 0.000 0.001 0.004 0.016 5.024 2 0.010 0.020 0.051 0.103 0.211 7.378 9.348 0.072 0.115 0.216 0.352 0.584 4 0.207 0.297 0.484 0.711 1.064 11.143 0.412 0.554 0.831 1.145 1.610 12.833 6 0.676 0.872 1.237 1.635 2.204 14.449 2.833 7 0.989 1.239 1.690 2.167 16.013 8 1.344 1.646 2.180 2.733 3.490 17.535 1.735 2.088 2.700 3.325 4.168 19.023 3.247 10 2.156 2.558 3.940 4.865 20.483 11 2,603 3.053 3.816 4.575 5.578 21.920 23.337 3.571 12 3.074 4.404 5.226 6.304 5.892 13 3.565 4.107 5.009 7.042 24.736 14 4.075 4.660 5.629 6.571 7.790 26.119 15 4.601 5.229 6.262 7.261 8.547 27.488 16 5.142 5.812 6,908 7.962 9.312 28.845 17 G408 7564 8.072 24.769 27.587 30.191

$\chi ^{2}=27.204$ fall between 24.769 and 27.587 , corresponding right tail area is between 0.05 to 0.10

Thus 0.05 < p-value < 0.10