Question

a. The drying time of a type of paint under specified test condition is known to be normally distributed with an assumed mean value of 75 and a standard deviation of 9. Test the hypothesis H0: µ=75 vs. Ha: µ<75 from a sample of 25 observations. For a test procedure with α= 0.002 what is the probability of making a type 2 error when the true mean is 70, ß(70) (to 4 decimals)?

b. Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorable, so a potential customer has decided to go ahead with a purchase arrangement unless it can be conclusively demonstrated that the true average lifetime is smaller than what is advertised. A random sample of 50 bulbs was selected, and the sample mean was found to be 738.44 with a standard deviation of 38.20 and a standard error on the mean of 5.40. What is the p-value for this test (to 4 decimals)?

Answer 1

a)

effect size is

ΕΡ μο - μ σ

d=5/9 = 0.5555

The probability of making a type Il error (B) is computed as follows: B = Pr(Fail to Reject a False Null Hypothesis Ho) -Mou) = Praze o/n = Praze X - μ-μ- μo o/n u) X - = = Pr(ze u - Mo + o/no/vn A) has a standard normal distribution. Therefore, the But, if the true population mean is u, then following is obtained: д- До B = Pr(ze <Z+ on = = Prazc H - Mo 0/vn <Z = Pr(ze + dyn<<Zc - dun) Observe that we have assumed without loss of generality that ji – Mo <0. Now, we plug the values we know to get:

B = Pr(-2.88 +0.5555 x 25 <2) = Pr(-0.1025 < Z) = 0.5408

we are allowed to solve one question only.