Question

equal" at the 5 percent siguificance level. 2. Still considering the data from "timetofail": redo the test, this time NOT assuming normality of the data. Use a non-parametric test. 1 type 1 type2 3.19 4.26 3.03 5.53 5.6 9.3 9.92 12.51 12.95 15.21 4.47 4.53 4.67 4.69 12.78 6.79 16.049.37 12.75 10 16.84 12 1. Consider the spreadsheet "timetofail". This contains data on the time to fail of a certain type of bolt that receives a type of treatment. The data is NOT paired data. If you assume normality of the data and assume equal variances, test the null hypothesis "the times to fail are equal" versus "not equal" at the 5 percent significance level. 2. Still considering the data from "timetofail": redo the test, this time NOT assuming normality of the data. Use a non-parametric test

Answer 1

1.

We find the mean and standard deviation of both the samples as:

For Type 1,

The sample size is n- 10. The provided sample data along with the data required to compute the sample mean X and sample variance s are shown in the table below: 3.03 9.1809 5.53 30.5809 5.6 31.36 9.3 86.49 9.92 98.4064 12.51 156.5001 12.95 167.7025 15.21 231.3441 16.04 257.2816 283.5856 16.84 106.93 1352.432 Sum-

The sample mean X is computed as follows: 106.93 = 10,693 8 Also, the sample variance s is 106.932 1352.432 23.226 10 --) 10- 1 n- 1 Therefore, the sample sandartd deviation s is /g? = V23.226 4.819 s

Ý, 10.693

Si-4.819

For Type 2,

The sample size is n- 10. The provided sample data along with the data required to compute the sample mean X and sample variance s are shown in the table below: 10.1761 3.19 18.1476 4.26 4.47 19.9809 20.5209 4.53 21.8089 4.67 4.69 21.9961 12.78 163.3284 6.79 46.1041 9.37 87.7969 12.75 162.5625 Sum- 67.5 572.422

The sample mean X is computed as follows: = 6.75 i16.75 TL Also, the sample variance s2 is 12.977 n-1 TL Therefore, the sample sandartd deviation s is 8-ve = V12.977 = 3.602

$\bar{X_2} =6.75$

Ss 3.602

Now, we conduct T-test for Means as :

The provided sample means are shown below: X 10.693 X 6.75 Also, the provided sample standard deviations are.: S1 4.819 82 3.602 and the sample sizes are n 10 and n2 10 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. (V) ke Based on the information provided, the significance level is α-0.05, and the degrees of freedom are df 18. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal

Hence, it is found that the critical value for this two-tailed test is t 2.101, for a 0.05 and df- 18. The rejection region for this two-tailed test is R = {t: t> 2.101} (3)_Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: n1+n2-2 2 10.693- 6.75 2.072 (10-1)4.8192+(10-1)3.60221 10+10-2 (4) Decision about the null hypothesis Since it is observed thatt2.072t 2.101, it is then concluded that the null hypothesis is not rejected Using the P-value approach: The p-value is p 0.0529, and since p-0.0529 2 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μι is different than μ., at the 0.05 significance level.

Hence, there is not enough evidence to conclude that the time to fail of a two type of bolts is different at 5% level of significance.

2.

Here, we conduct Wilcoxon Rank-Sum Test as :

First, we put both samples together and organize it in ascending order, which is shown in the table below: Sample Value 3.03 3.19 4.26 4.47 4.53 4.67 4.69 5.53 5.6 6.79 9.3 9.37 9.92 12.51

12.75 12.78 12.95 15.21 6.04 6.84 Now, that the values that are in ascending order are assigned ranks to them, taking care of assigning the average rank to values with rank ties Rank (Adjusted for Rank Value Sample 3.03 3.19 4.26 4.47 4.53 4.67 4.69

5.53 5.6 6.79 10 10 9.3 9.37 12 12 9.92 13 13 12.51 12.75 15 15 16 12.78 16 17 17 12.95 18 15.21 18 19 16.04 19 20 20 16.84 The sum of ranks for sample 1 is R1 18+9+1113+ 14+17 +18+19+20- 130

and the sum of ranks of sample 2 is R2 2+3+4+5+6+7+10+12 +15 +16 80 Hence, the test statistic is RR1 130. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested:

Ho : Median(Difference)0

H, : Median(Difference) 0

The critical value for the signficance level provided and the type of tail is Rc hypothesis is rejected if R 〈 52. 52, and the null (3) Decision about the null hypothesis Since in this case R 130 > 52, there is not enough evidence to claim that the population median of differences is different than 0, at the 0.05 significance level.

Hence, there is not enough evidence to conclude that the time to fail of a two type of bolts is different at 5% level of significance.