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

good luck!

cutie~

the correct answer is 5 6 and 7