Question

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).

Answer 1

Answer:

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.

Answer 2

Answer:

Step-by-step explanation:

Hello!

The manufacturer suspects that the time a part for cell phones is allowed to cool in the mold influences the occurrence of a defect in the finished housing.

The claim is that the longer the part is left to cool down, the fewer defects appear.

To test this claim, two random samples of 20 parts each were taken, one was left to cool down 10sec and the other was left to cool down for 20sec. After the cooling time, each one was visually inspected and scored based on their appearance from 1 (lowest) to 10 (highest).

Resulting:

X₁: Score given to a manufactured part after a cooling time of 10 seconds.

n₁= 20 parts

X[bar]₁= 3.35

S₁= 2.01

X₂: Score given to a manufactured part after a cooling time of 20 seconds.

n₂= 20 parts

X[bar]₂= 6.50

S₂= 1.54

Assuming both variables have a normal distribution, and the population variances are equal, the statistic to use is a student t for two independent samples with pooled sample standard deviation.

If the longer the cooling time, the fewer defects appear in the part, then the mean score will be greater for the sample with higher cooling time.

The statistical hypotheses are:

H₀: μ₁ ≥ μ₂

H₁: μ₁ < μ₂

α: 0.05

$t_{H_0}= \frac{((X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_1} } } ~~t_{n_1+n_2-2}$

$Sa^2= \frac{(n_1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2}$

$Sa^2= \frac{19*4.04+19*2.37}{20+20-2} = 3.81$

Sa= 1.95

$t_{H_0}= \frac{(3.35-6.50)-0}{1.95\sqrt{\frac{1}{20} +\frac{1}{20} } } = -5.108$

This test is one tailed to the left and so is the p-value, you can reach the value as follow:

p-value: $P(t_{38}\leq -5.108)$ ≅ 0.00001

p-value < 0.00001

The p.value is less than the significance level, the decision is to reject the null hypothesis.

Using a level of significance of 5% there is enough statistical evidence to reject the null hypothesis, so you can say that the average inspection scores of the parts that cooled down for 10 seconds are less than the average inspections scores of the parts that cooled down for 20 seconds, which means that the longer the cooling time, the fewer defects appear in the cellphone housing.

*-*-*

95% CI for the difference of means μ₁ - μ₂

The formula for the confidence interval is:

[(X[bar]₁-X[bar]₂) ± $t_{n_1+n_2-2;1-\alpha /2} * (Sa \sqrt{\frac{1}{n_1} +\frac{1}{n_2} } )$]

$t_{n_1+n_2-2;1-\alpha /2} = t_{38;0.975}= 2.024$

[(3.35-6.50)±2.024*(1.95*$\sqrt{\frac{1}{20} +\frac{1}{20} }$)]

[-4.40;-1.90]

With a confidence level of 95%, you'd expect that the true value of the difference between the inspection scores of the parts that cooled down for 10 seconds and inspections scores of the parts that cooled down for 20 seconds will be included in the interval [-4.40;-1.90].

I hope it helps!