Material science engineers would like to determine if there is a significant dif

Question

Material science engineers would like to determine if there is a significant difference in average strength between two different heat treatment procedures for an Aluminum alloy. A tensile test is then conducted to determine the material strength. For both process, 15 specimens were tested, and found that heat treatment 1 samples have an average yield strength of 280.5 MPa with a standard deviation of 14.08, and heat treatment 2 samples have an average yield strength of 308.2 MPa with a standard deviation of 12.67. Assume that the observations come from normal populations with different variance.

Question 1:

What kind of hypothesis test would you use?

Question 2:

Question 3:

What is the alternative hypothesis?

Question 4:

Would you use a z-test or a t-test?

An upper-tail t-test

Question 5:

What is the value of the test statistic you obtain?

Question 6:

What is the critical value for a significance level of 0.05? (provide the negative critical value if it is a two-sided hypothesis test)

Question 7:

What would you conclude from this hypothesis test?

Since the test statistic falls in the acceptance region, we should accept H0, and conclude that the average tensile strengths from the two heat treatments are not statistically different.

Since the test statistic falls in the rejection region, we should reject H0, and conclude that the average tensile strengths from the two heat treatments are statistically different.

Since the test statistic falls in the acceptance region, we should accept H1, and conclude that the average tensile strengths from the two heat treatments are statistically different.

Since the test statistic falls in the do not reject region, we should not reject H0, and conclude that the average tensile strengths from the two heat treatments are not statistically different.

Question 8:

Assume that you conducted a similar hypothesis test, and found the test statistic is -2.15, with a degree of freedom of 27. At what significance level would you reject the null hypothesis?

Explanation / Answer

Question-1:

As 15 specimen are tested independently in each treatment and so total sample=30, so unpaired two mean test:

An unpaired-two-population mean hypothesis test

Question-2:

As we have to test significant difference so null hypothesis is second option

Question-3:

First option

Question-4:

As sample sizes are less than 30, so and alternative is two tailed,so

A two-tail t-test

Question-5:

Test statistic t= (xbar1-xbar2)/sqrt(var1/n1+var2/n2)

=(280.5-308.2)/sqrt(14.08*14.08/15+12.6*12.67/15)

=-5.67

Question-6:

We have C=(var1/n1)/(var1/n1+Var2/n2) = (14.08*14.08/15)/(14.08*14.08/15+12.6*12.67/15)=0.55

Degree of freedom =((n1-1)*(n2-1))/((n2-1)c*c+(1-c)*(1-c)(n1-1))

=(15-1)*(15-1)/((15-1)*0.55*0.55+(1-0.55)*(1-0.55)*(15-1))

=27.72

=27

Critical value at 5%level is t=- 2.0518, 2.0518

Question-7:

Since the test statistic falls in the rejection region, we should reject H0, and conclude that the average tensile strengths from the two heat treatments are statistically different.

Question-8:

p-value (two tailed)= 0.0407

So, we reject the null hypothesis at =0.05,0.20,0.10

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