The output voltage of a certain electric circuit is claimed to be 130 volts. The

Question

The output voltage of a certain electric circuit is claimed to be 130 volts. The population standard deviation is known to be sigma = 3.0 volts. A sample of 40 readings on the voltage of this circuit gave a sample mean of 128.6 volts. We want to test the null hypothesis that the mean output voltage is 130 volts against the alternative that it is less than 130 volts, using two significance levels: (1) alpha = 0.05 and (2) alpha = 0.01. Suppose that the true population mean is 129.6 volts. Please answer the following questions with explanations a. Which of (1) and (2) gives the smaller type I error? b. Which of (1) and (2) gives the smaller type II error? c. Do we need a smaller or larger sample size if we want to reduce the type II errors for both (1) and (2)? d. If the true population mean output voltage is in fact 128.6 volts, how will the type I errors of (1) and (2) change? How will the powers of (1) and (2) change?

Explanation / Answer

a. Output voltage = 130 volts

Population standard deviation = 3.0 volts

Sample size = 40

sample mean = 128.6 volts

Null Hypothesis : H0 : voltage = 130 volts

Alternative Hypothesis : Ha : voltage < 130 volts

= 0.05 and = 0.01

True population mean = 129.6 volts

a. Type I error is obviously small for alpha = 0.01 as it is less than alpha = 0.05. The type I error means rejecting the null hypothesis even if it is correct. so here alpha is the type I error so for less alpha, type I error will be less.

Rejection Region for alpha = 0.01 is (xbar < z) where z = 130 - t0.01, 39 (/n) = 130 - 2.426 * (3/ 40) = 130 - 1.15 = 128.85 volts

Rejection Region for alpha = 0.05 is (xbar < Z) where Z = 130 - t0.05 39 (/n) = 130 - 1.685 * (3/40) = 130 - 0.8 = 129.2 volts

so we can say that (1) give smaller type I error.

(b) Type II error is to accept the null hypothesis even if it is false. so in both cases we will accept the null hypothesis even if it is false.

So Type II error when alpha = 0.01 is = Pr(xbar >= 128.85 ; 129.6; 3/40)

Here Z - value = ( 128.85 - 129.6)/ (3/40) = -0.75/ 0.4743 = - 1.58

so Pr(xbar >= 128.85 ; 129.6; 3/40) = 1- (-1.58) = 1 - 0.0571 = 0.9429

Type II error when alpha = 0.05 is = Pr(xbar >= 1292 ; 129.6; 3/40)

Here Z - value = ( 129.2 - 129.6)/ (3/40) = -0.40/ 0.4743 = - 0.844

so Pr(xbar >= 129.2 ; 129.6; 3/40) = 1- (-0.84) = 1 - 0.2000 = 0.800

so type II error for (1) is more than from (2) .

(c) We need a small sample size to reduce the type II error. as square root of sample size is inversly proportion to prbability of type II error.

(d) If we change true mean to 128.6 Volts, type I errors as (1) and (2) will be the same because change in true mean will not effect the value of type I error.

Type II error when alpha = 0.01 is = Pr(xbar >= 128.85 ; 128.6; 3/40)

Here Z - value = ( 128.85 - 128.6)/ (3/40) =0.25/ 0.4743 = 0.527

so Pr(xbar >= 128.85 ; 129.6; 3/40) = 1- (0.527) = 1 - 0.7019 = 0.2981

Power = 1 - 0.2981 = 0.7019

Type II error when alpha = 0.05 is = Pr(xbar >= 129.2 ; 128.6; 3/40)

Here Z - value = ( 129.2 - 128.6)/ (3/40) = 0.60/ 0.4743 = 1.265

so Pr(xbar >= 129.2 ; 129.6; 3/40) = 1- (1.265) = 1 - 0.8961 = 0.1039

Power = 1- 0.1039 = 0.8961

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