Question

7.97 Suppose you want to test Ho: u = 500 against He: p > 500 using a = .05. The population in question is normally dis- tributed with standard deviation 100. A random sample of size n = 25 will be used. a. Sketch the sampling distribution of x assuming that Ho is true. b. Find the value of xo, that value of x above which the null hypothesis will be rejected. Indicate the rejection region on your graph of part a. Shade the area above the rejection region and label it a. c. On your graph of part a, sketch the sampling distribu- tion of x if u = 550. Shade the area under this distri- bution that corresponds to the probability that x falls in the nonrejection region when u = 550. Label this area ß. d. Find ß. e. Compute the power of this test for detecting the alter- native Ha: M = 550. 7.98 Refer to Exercise 7.97. a. If u = 575 instead of 550, what is the probability that the hypothesis test will incorrectly fail to reject H? That is, what is B? b. If u = 575, what is the probability that the test will correctly reject the null hypothesis? That is, what is the power of the test? c. Compare ß and the power of the test when u = 575 to the values you obtained in Exercise 7.97 for j = 550. Explain the differences.

Answer 1

7.97)

H₀ : µ = 500 vs H_a : µ > 500  

Given :    σ = 100 , n = 25 , $\sigma _{\bar{x}}$ = = 20

ulo

a)

40 460 480 500 520 540 560

b) This is right tail test , α = 0.05 , Z_0.05 = 1.645

₀ = z*$\sigma _{\bar{x}}$  + µ = (1.645*20 ) + 500

₀ = 532.9

Rejection region = If is greater than 532.9 , then we reject H₀

Fail to reject HO Rejection region of HO + 500 532.9

c)

500 532. 9 550

d) $\beta$ = P( Fail to reject H₀ Given H₀ is false )

= P( ≤ 532.9 | µ = 550 )

=  

p(* -Mi s 532.9 – 550 p( $ 0 20

= P( z ≤ -0.86 )

$\beta$ = 0.1949

e) Power of the test = 1 - $\beta$ = 1 - 0.1949

Power of the test = 0.8051