A bearing used in an automotive application is suppose to have a nominal inside

Question

A bearing used in an automotive application is suppose to have a nominal inside diameter of 1.5 inches. A random sample of 25 bearings is selected and the average inside diameter of these bearings is 1.4975 inches. Bearing diameter is known to be normally distributed with standard deviation inch. (a) Test the hypotheses H0: mu equals 1.5 versus H1: mu not-equals 1.5 using alpha equals 0.01 (b) What is the P-value for the test in part (a)? (c) Compute the power of the test if the true mean diameter is 1.495 inches. (d) What sample size would be required to detect a true mean diameter as low as 1.495 inches if we wanted the power of the test to be at least 0.86?

Explanation / Answer

A bearing used in an automotive application is suppose to have a nominal inside diameter of 1.5 inches. A random sample of 25 bearings is selected and the average inside diameter of these bearings is 1.4975 inches. Bearing diameter is known to be normally distributed with standard deviation inch.

(a) Test the hypotheses H0: mu equals 1.5 versus H1: mu not-equals 1.5 using alpha equals 0.01

Minitab used

One-Sample Z

Test of = 1.5 vs 1.5

The assumed standard deviation = 0.01

N     Mean SE Mean        99% CI            Z      P

25 1.49750 0.00200 (1.49235, 1.50265) -1.25 0.211

Calculated z=-1.25 not falls in the rejection region ( -2.576, 2.576).

The null hypothesis is not rejected.

We conclude that bearing have inside diameter of 1.5 inches.

(b) What is the P-value for the test in part (a)?

P=0.211

(c) Compute the power of the test if the true mean diameter is 1.495 inches.

Power and Sample Size

1-Sample Z Test

Testing mean = null (versus null)

Calculating power for mean = null + difference

= 0.01 Assumed standard deviation = 0.01

            Sample

Difference   Size      Power

    0.0025      25 0.0925133

Power =0.0925

(d) What sample size would be required to detect a true mean diameter as low as 1.495 inches if we wanted the power of the test to be at least 0.86?

Power and Sample Size

1-Sample Z Test

Testing mean = null (versus null)

Calculating power for mean = null + difference

= 0.01 Assumed standard deviation = 0.01

            Sample Target

Difference    Size   Power Actual Power

    0.0025     214    0.86      0.860230

Sample size required =214

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