By NEAS - 7/2/2024 4:51:05 PM
MS Module 4 type 1 and type 2 errors practice exam questions
(The attached PDF file has better formatting.)
A population has a normal distribution with a mean μ0 of 60 and a standard deviation of 7.3
One group from this population has been treated to reduce its mean; we assume it is still normally distributed with the same standard deviation. A sample of size 22 from this treated group has a sample mean of and a true mean of μʹ.
● The null hypothesis is H0: μʹ = μ0.
● The one-sided alternative hypothesis is Ha: μʹ < μ0.
We reject the null hypothesis if ≤ 58.8
Question 4.1: Standard deviation of sample mean
What is the standard deviation of the sample mean?
Answer 4.1: 7.3 / 220.5 = 1.556
(standard deviation of the sample mean = standard deviation / (number of observations in sample)0.5 )
Question 4.2: z statistic
What is the z statistic value to test the null hypothesis?
Answer 4.2: (58.8 – 60) / 1.556 = -0.771
(the z statistic value to test the null hypothesis = (sample mean – mean assumed in null hypothesis (μ0) ) / standard deviation of the sample mean)
Question 4.3: Probability of Type I error
What is the probability of a Type I error for this one-sided (lower-tailed) test?
Answer 4.3: Φ(–0.771) = 0.2203
Interpolating in the statistical tables:
Φ(0.77) = 0.7794
Φ(0.78) = 0.7823
Φ(–0.771) = 1 – ( (0.771 – 0.77) × 0.7823 + (0.78 – 0.771) × 0.7794) / (0.78 – 0.77) = 0.2203
Question 4.4: Probability of Type II error
If the true mean of the sample μʹ is 59.3 , what is the probability of a Type II error for this test?
Answer 4.4: (59.3 – 58.8) / 1.556 = 0.3213, so Φ(59.3 – 58.8) / 1.556 = Φ(0.3213) = 0.6260
Interpolating in the statistical tables:
Φ(0.32) = 0.6255
Φ(0.33) = 0.6293
Φ(0.3213) = ( (0.3213 – 0.32) × 0.6293 + (0.33 – 0.3213) × 0.6255) / (0.33 – 0.32) = 0.6260
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