Question

 Analysis of a random sample consisting of m = 20 specimens of cold-rolled steel to determine yield strengths resulted in a sample average strength of ¯x = 29.8 ksi. A second random sample of n = 25 two-sided galvanized steel specimens gave a sample average strength of ¯y = 34.7 ksi. Assuming that the two yield-strength distributions are normal with σ1 = 4.0 and σ2 = 5.0.

(a) Find the 99% confidence interval of the difference µ1 − µ2 between the population means of the two normal distributions. 

(b) Does the data indicate that the corresponding true average yield strengths µ1 and µ2 are different? Carry out a test at significance level α = 0.01. Write down the test statistic, its value, the rejection region, and the scientific conclusion as your answer.

Answer 1

To find the 99% confidence interval for the difference between the population means, µ1 - µ2, you can use the following formula:

Confidence Interval = (¯x - ¯y) ± Z_(α/2) * √[(σ1^2/m) + (σ2^2/n)]

Where:

• ¯x = sample average strength for cold-rolled steel = 29.8 ksi

• ¯y = sample average strength for galvanized steel = 34.7 ksi

• σ1 = standard deviation for cold-rolled steel = 4.0 ksi

• σ2 = standard deviation for galvanized steel = 5.0 ksi

• m = size of the cold-rolled steel sample = 20

• n = size of the galvanized steel sample = 25

• Z_(α/2) is the critical value for a 99% confidence interval. You can find this value using a standard normal distribution table or calculator. For a 99% confidence interval, α/2 = 0.005, so Z_(0.005) ≈ 2.576.

Now, let's calculate the confidence interval:

Confidence Interval = (29.8 - 34.7) ± 2.576 * √[(4^2/20) + (5^2/25)] Confidence Interval = -4.9 ± 2.576 * √[(0.8) + (1.0)] Confidence Interval = -4.9 ± 2.576 * √(1.8)

Now, calculate the values:

Confidence Interval = -4.9 ± 2.576 * 1.34164

Calculate the upper and lower bounds of the confidence interval:

Lower Bound: -4.9 - 2.576 * 1.34164 ≈ -8.702 Upper Bound: -4.9 + 2.576 * 1.34164 ≈ -1.098

So, the 99% confidence interval for the difference between the population means (µ1 - µ2) is approximately (-8.702, -1.098) ksi.

(b) To test if the true average yield strengths µ1 and µ2 are different, you can perform a two-sample t-test at a significance level α = 0.01. The null hypothesis (H0) is that there is no difference between the two population means, and the alternative hypothesis (Ha) is that there is a difference.

The test statistic for a two-sample t-test is given by:

t = (¯x - ¯y) / √[(s1^2/m) + (s2^2/n)]

Where:

• ¯x and ¯y are the sample means.

• s1 and s2 are the sample standard deviations.

• m and n are the sample sizes.

We already have the sample means, standard deviations, and sample sizes. So:

t = (29.8 - 34.7) / √[(4^2/20) + (5^2/25)] t = (-4.9) / √[(0.8) + (1.0)] t ≈ -4.9 / √1.8 t ≈ -4.9 / 1.34164 ≈ -3.65

The degrees of freedom (df) for this test is df = (m + n - 2) = (20 + 25 - 2) = 43.

At a significance level α = 0.01, you can find the critical t-value from a t-distribution table with df = 43. The critical t-value for a two-tailed test at α/2 = 0.005 is approximately ±2.694.

Since the calculated t-value (-3.65) is less than the critical t-value (-2.694), we can reject the null hypothesis.

Conclusion: At the 0.01 significance level, there is enough evidence to conclude that the true average yield strengths µ1 and µ2 are different.