An experiment conducted for the specific purpose of comparing the compressive st

Question

An experiment conducted for the specific purpose of comparing the compressive strength of structural components for marine hoisting vehicles yielded the following coded results: a) Assume that the true standard deviations are known and equal with those in row 3 of the table. Test the hypothesis that the mean compressive strength of rails manufactured by A is not significantly different from that of B. Let alpha = 0.05. State any assumptions you need to make. b) Construct a 95% confidence interval on the mean difference in compressive strength (true standard deviations known and equal with the values in row 3). c) Assume that only the sample standard deviations are known (row 4 in the table). Test the hypothesis that the mean compressive strength of rails manufactured by A is not significantly different from that of B. Let alpha = 0.05. State any assumptions you need to make. Compare the result with part (a) and explain the difference. d) Construct a 95% confidence interval

Explanation / Answer

Question a)

The population distribution for Manufacturer A and Manufacturer A for are normally distributed.

Here the population standard deviations are known so we will use z-test.

Null and Alternative Hypothesis:

Ho: µA = µB

Ha: µA not equal to µB

Level of significance is .05

z = (XA bar – XB bar )/sqrt ((sigmaA^2/nA) + (sigmaB^2/nB))

XA bar = 9890

XB bar = 10120

sigma A = 220

sigma B = 310

nA = 15

nB = 10

z = (9890 - 10120)/sqrt((220^2/15)+(310^2/10))

= -2.03

The critical values at 5% level of significance are given as -1.96 and +1.96 (from normal table).

The test statistics -2.03 is less than -1.96, test statistics falls in critical region; we reject the null hypothesis.

At 5% level of significance there is sufficient evidence to conclude that the mean compressive strength of rails manufactured by A is significantly different from that of B.

Question b)

Here also we use z distribution as population standard deviations are known here.

Confidence Interval:

(X A bar – XB bar) (-/+) E (XA – XB)

E (XA-XB) = zc * sqrt ((sigmaA^2/nA) + (sigmaB^2/nB))

                  = 1.96* sqrt((220^2/15)+(310^2/10))

                  = 222.07

(XA bar – XB bar) = 9890 – 10120 = -230

-230 (-/+) 222.07

-452.07 and -7.93

95% confidence interval is (-452.07 and -7.93)

Question c)

The population distribution for Manufacturer A and Manufacturer A for are normally distributed. The population variances are equal for the two populations.

Here population standard deviations are not given so we use t-test.

Null and Alternative Hypothesis:

Ho: µA = µB

Ha: µA not equal to µB

Level of significance is .05

z = (XA bar – XB bar )/SE (XA-XB)

SE (XA – XB) = sqrt((nA-1)*sigmaA^2+(nB-1)*sigmaB^2)/(nA1+nB-2) * sqrt(1/nA+1/nB)

XA bar = 9890

XB bar = 10120

sigma A = 220

sigma B = 310

nA = 15

nB = 10

t= (9890 - 10120)/(sqrt(((15-1)*220^2+(10-1)*(310^2))/(15+10-2))*sqrt((1/15)+(1/10))

= -2.175

Here degrees of freedom is 23 (10+15-2) and the level of significance is .05

The critical values at 5% level of significance are given as -2.069 and +2.069 (from t table).

The test statistics -2.175 is less than -2.069 test statistics falls in critical region; we reject the null hypothesis.

At 5% level of significance there is sufficient evidence to conclude that the mean compressive strength of rails manufactured by A is significantly different from that of B.

In part ( a ) also we got the same conclusion as there is a significant difference between the mean compressive strength of rails manufactured by A and B.

Question d)

Here also we use t distribution as population standard deviations are unknown here.

Confidence Interval:

(X A bar – XB bar) (-/+) E (XA – XB)

E (XA-XB) = tc * sqrt((nA-1)*sigmaA^2+(nB-1)*sigmaB^2)/(nA1+nB-2) * sqrt(1/nA+1/nB)

                  = 2.069 * (sqrt(((15-1)*220^2+(10-1)*(310^2))/(15+10-2))*sqrt((1/15)+(1/10))                  =

                 = 218.74

(XA bar – XB bar) = 9890 – 10120 = -230

-230 (-/+) 218.74

-448.74 and -11.26

The 95% confidence interval is (-448.74 and -11.26).

We find the width (upper bound – lower bound) of the confidence interval got decreased in case of part ( d ) compare to in part ( b ).

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