Mathematical Statistics with Applications
Mathematical Statistics with Applications
7th Edition
ISBN: 9781133384380
Author: Dennis Wackerly; William Mendenhall; Richard L. Scheaffer
Publisher: Cengage Learning US
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Chapter 13.7, Problem 30E

a.

To determine

Check whether there is sufficient evidence to indicate a difference in the mean wear among the four treatments using α=0.05 significance.

a.

Expert Solution
Check Mark

Answer to Problem 30E

There is sufficient evidence that at least one mean is different.

Explanation of Solution

The null and alternative hypotheses are as follows:

Null hypothesis: There is no significant difference in mean wear among the four treatments.

Alternative hypothesis: There is significant difference in mean wear among the four treatments.

The sample means are as follows:

Y¯1.=j=1n1y1jn1=9.16+13.29+12.07+11.97+13.31+12.32+11.787=11.98571

Y¯2.=j=1n2y2jn2=11.95+15.15+14.75+14.79+15.48+13.47+13.067=14.09286

Y¯3.=j=1n3y3jn1=11.47+9.54+11.26+13.66+11.18+15.03+14.867=12.42857

Y¯4.=j=1n4y4jn2=11.35+8.73+10+9.75+11.71+12.45+12.387=10.91

Y¯=Y¯1.+Y¯2.+Y¯3.+Y¯4.4=12.35429

Total sum of square is calculated as follows:

SSE=i=14j=17(YijY¯i.)2=[(9.1611.98571)2+(13.2911.98571)2+(12.0711.98571)2+(11.9711.98571)2+(13.3111.98571)2+(12.3211.98571)2+(11.7811.98571)2+(11.9514.09286)2+(15.1514.09286)2+(14.7514.09286)2+(14.7914.09286)2+(15.4814.09286)2+(13.4714.09286)2+(13.0614.09286)2+(11.4712.42857)2+(9.5412.42857)2+(11.2612.42857)2+(13.6612.42857)2+(11.1812.42857)2+(15.0312.42857)2+(14.8612.42857)2+(11.3510.91)2+(8.7310.91)2+(1010.91)2+(9.7510.91)2+(11.7110.91)2+(12.4510.91)2+(12.3810.91)2+]=60.282

Sum of square for treatments is calculated as follows:

SST=i=14ni(Y¯i.Y¯)2=0.951+21.158+0.039+14.602=36.750

The mean squares for the treatments are as follows:

MST=SSTk1=36.7541=12.25

The mean square for the error is as follows:

MSE=SSEn4=60.282284=2.5118

The F-value is as follows:

F=MSTMSE=12.252.5118=4.877

Critical value:

The critical value is obtained from F table with 3 degrees of freedom for the numerator and 24 for the denominator.

Step-by-step procedure to obtain F-value using Table 7 of Appendix 3:

  • Locate the value 3 and α=0.05 in the right column of Table 7 of Appendix 3.
  • Go through the row corresponding to the value 3 and column corresponding to the value 24 of the Table 7.
  • Locate the value corresponding to (3, 24).

The required critical value for F-distribution with the level of significance 0.05, df1=3 and df2=24 is 3.01.

Rejection rule:

If FF-critical, reject H0. Otherwise fail to reject H0.

Conclusion:

Here, the F-test statistic is 4.877.

The test statistic (=4.877) is greater than the critical value (=3.01).

Therefore, the null hypothesis is rejected.

Thus, there is sufficient evidence to conclude that at least one mean is different from others.

b.

To determine

Construct a 99% confidence interval for the difference in haze increase between treatments B and C.

b.

Expert Solution
Check Mark

Answer to Problem 30E

The 99% confidence interval for the difference in haze increase between treatments B and C is (0.70521,4.03379).

Explanation of Solution

Y¯B=14.09286,Y¯C=12.42857, MSE =2.5118=S2,nk=24,nA=7 and nB=7.

The confidence interval for the difference in the means for treatments i and i’ is as follows:

CI=(Y¯i.Y¯i'.±tα2S1n1+1n2)

Here, tα2 is based on the (nk) degrees of freedom.

From “Appendix 3, Table 5 Percentage Points of the t Distributions”, the critical value under the level of significance 0.01(=α) and 24 degrees of freedom is 2.797.

Substitute the values in the confidence interval and the 99% confidence interval for the difference in haze increase between treatments B and C:

CI=(Y¯i.Y¯i'.±tα2S1n1+1n2)=(14.0928612.42857±(2.797)2.511817+17)=(1.66429±2.3695)=(0.70521,4.03379)

Thus, the 99% confidence interval for the difference in haze increase between treatments B and C is (0.70521,4.03379).

c.

To determine

Construct a 90% confidence interval for the mean wear for lenses receiving treatment A.

c.

Expert Solution
Check Mark

Answer to Problem 30E

The 90% confidence interval for the mean wear for lenses receiving treatment A is (10.961,13.011).

Explanation of Solution

For mean treatment A, Y¯A=11.98571,MSE =2.5118=S2,nk=24,nA=7.

The confidence interval for the mean treatment i is as follows:

CI=(Y¯i.±tα2Sni)

Here, tα2 is based on the (nk) degrees of freedom.

From “Appendix 3, Table 5 Percentage Points of the t Distributions”, the critical value under the level of significance 0.05(=α) and 24 degrees of freedom is 1.711.

Substitute the values in the confidence interval and the 90% confidence interval for the mean wear for lenses receiving treatment A:

CI=(Y¯i.±tα2Sni)=(11.98571±(1.711)2.51187)=(10.961,13.011)

Thus, the 90% confidence interval for the mean wear for lenses receiving treatment A is (10.961,13.011).

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Chapter 13 Solutions

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