A man has a few drinks one night at the bar and is deciding whether to drive hom

Question

A man has a few drinks one night at the bar and is deciding whether to drive home at the end of the night. He is essentially testing the hypotheses H_0: My blood-alcohol level is below the legal limit vs. H_a: My blood-alcohol level is above the legal limit Explain what it would mean in the context of this example to make a Type I error and a Type II error. Explain the potential consequences of each type of error. (b) The strengths of prestressing wires manufactured by a steel company have a mean of 2000 N and a standard deviation of 100 N. By employing a new manufacturing technique, the company claims that the mean strength will be increased. To verify this claim, a builder will test a random sample of 36 wires produced by the new process and will conduct a hypothesis test of H_0: mu = 2000 vs. H_a: mu > 2000 at the 10% level of significance. What would be the power of the test if the true mean strength of wires produced by the new process was 2050 N?

Explanation / Answer

(A) Here Type I error is rejecting the null hypothesis even if it is true that means even he drank below the legal limit, he will consider himself illegible for driving at night.

Here in this case. Type II error is to not reject the null hypothesis even it is false that means even he drank above the legal limit, he will consider himself illegible for driving at night.

Type I error will not have very much negative consequences as he can take a cab or help of her friend to reach home but type II error will have a very negative consequences.

B) Number of samlpes = 36

H0: µ = 2000 and H1: µ > 2000

level of significance alpha = 0.10

Here True mean = 2050 N

so Here type II error will occur when we will not reject the null hypothesis even if it is false.

Here The region of acceptance of null hypothesis = µH + Z0.1 (/n) = 2000 + 1.28 * (100/36)

= 2000 + 1.28 * (100/6) = 2021.33

so region of acceptance of null hypothesis is xbar< 2021.33

so we can commit the type II error when Pr( xbar< 2021.33 ; µT ; /n)

Pr( xbar< 2021.33 ; 2050; 16.67) = ?

HEre Z- value = ( 2021.33 - 2050)/ 16.67 = -1.72

so Value of probability from Z - table for z = -1.72 is

Pr( xbar< 2021.33 ; 2050; 16.67) = 0.0427

Power of the test = 1 - 0.0427 = 0.9573

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