The diameters of aluminum alloy rods produced on an extrusion machine are known to have a standard deviation of 0.0001 in. A random sample of 25 rods has an average diameter of 0.5046 in.

a) Test the hypothesis that mean rod diameter is 0.5025 in. Assume two-sided alternative and significance level of 0.05.

b) Find the p-value for test in part (a).

c) Construct a 95% two-sided confidence interval on the mean rod diameter.

a) Hypothesis mean differs from 0.5025. Rejected

b) P-value = 0

c) 0.50456 < u < 0.50464

Step-by-step explanation:

Given:

- The standard deviation s. d = 0.0001 in

- The mean of the sample x = 0.5046 in

- Sample size n = 25

Find:

a) Test the hypothesis that mean rod. Assume two-sided alternative and significance level of 0.05.

b) Find the p-value for test in part

c) Construct a 95% two-sided confidence interval on the mean rod diameter.

Solution:

- The random variable X: the measurements for rod diameter follows a normal distribution:

X~ N (0.5025, 0.0001)

- Testing H_o: u_o = 0.5025, H_1 : u_o not equal 0.5025

- Compute the corresponding Z-score:

Z_o = (0.5046 - 0.5025) / 0.0001/sqrt (25)

Z_o = 105

- The Z score value with respect to significance level is:

Z_a/2 = Z_0.025 = 1.96

Hence, Z_o > Z_a/n

- The hypothesis H_o is rejected the mean differs from 0.5025.

- The corresponding P-Value is:

P-value = 2 * (1 - sig (Z_o)) = 2 * (1 - sig (105))

P-value = 2 * (1-1) = 0

- The confidence interval is:

(x - Z_a/2 (s. d/sqrt (n)) < u < (x + Z_a/2 (s. d/sqrt (n))

(0.5046 - 1.96 (0.0001/sqrt (25)) < u < (0.5046 + 1.96 (0.0001/sqrt (25))

0.50456 < u < 0.50464