Front housing for cell phones are manufactured in an injection molding process. The time the part is allowed to cool in the mold is thought to influence the occurrence of a defect in the finished housing. After manufacturing, the parts are inspected visually and assigned a score between 1 and 10, based on their appearance, with 10 being perfect, and 1 being completely defective.

An experiment was conducted using two cool-down times, 10 and 20 seconds. A random sample of 20 housings was collected for each cool-down time. The data is:

10s 20s
1 7
2 8
1 5
3 9
5 5
1 8
5 6
2 4
3 6
5 7
3 6
6 9
5 5
3 7
2 4
1 6
6 8
8 5
2 8
3 7
Is there evidence to support the claim that longer cool down results in fewer appearance defects? Write the hypotheses using mathematical notation, derive the test statistic and write the distribution of the test statistic under the null hypothesis, compute the p-value and draw your conclusions. Use a=0.05.

Find a 95% confidence interval for the difference between the means (between the 10-second cool down sample, and the 20-second sample).

A) P value = 0.000 and there is evidence that the longer cool-down time results in fewer appearance defects.

The test statistic is tabulated in explanations column.

B) confidence interval is µ1 - µ2 ≤ −2.196 ;

Thus, the 20 second cooling time gives a cosmetically better housing.

Step-by-step explanation:

A) Mean at 10 seconds = 67/20 = 3.35

Mean at 20 seconds = 130/20 = 6.5

From this, we calculate Standard deviation and standard error mean and tabulate as below;

Two-sample T for 10 seconds vs 20 seconds;

For 10 secs and N=20; mean =3.35; StDev=2.01 and SEMean= 0.45

For 20 secs and N= 20; mean = 6.50;

StDev = 1.54 and SEMean= 0.34

Estimate of Difference = mean at 10 seconds - mean at 20 seconds = 3.35 - 6.50 = - 3.15

Using the confidence limit calculator, we get;

95% upper bound for difference = -2.196

Also using the mean, standard deviation and degree of freedom,

T-Test of difference = 0 (vs <): T-Value = -5.57

P-Value = 0.000

DF(Degree of freedom)= (20-1) + (20-1) = 38

Both use Pooled StDev = (2.01 + 1.54)/2 = 3.55/2 = approximately 1.78

From the analysis shown below, there is evidence that the longer cool-down time results in fewer appearance defects.

B) From the tabulated output above,

µ1 - µ2 ≤ −2.196 .

This lower confidence bound is less than 0 and the two samples are different. Thus, the 20 second cooling time gives a cosmetically better housing.

the correct answer is d

hope this

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step-by-step explanation: