Question

Recall that "very satisfied" customers give the XYZ-Box video game system a rating that is at least 42. Suppose that the manufacturer of the XYZ-Box wishes to use the random sample of 63 satisfaction ratings to provide evidence supporting the claim that the mean composite satisfaction rating for the XYZ-Box exceeds 42.   

(a) Letting µ represent the mean composite satisfaction rating for the XYZ-Box, set up the null hypothesis H₀ and the alternative hypothesis H_a needed if we wish to attempt to provide evidence supporting the claim that µ exceeds 42.

H₀: µ (Click to select)≤≥≠><= 42 versus H_a: µ (Click to select)≤>=<≠≥ 42.

(b) The random sample of 63 satisfaction ratings yields a sample mean of x⎯⎯=42.860x¯=42.860. Assuming that σ equals 2.67, use critical values to test H₀ versus H_a at each of α = .10, .05, .01, and .001. (Round your answer z_.05 to 3 decimal places and other z-scores to 2 decimal places.)

z =      

Rejection points
z.10
z.05
z.01
z.001

Reject H₀ with α = (Click to select).01, .001.10.10, .05, .01.001 , but not with α =(Click to select).10.01, .001.001.10, .05, .01

(c) Using the information in part (b), calculate the p-value and use it to test H₀ versus H_a at each of α = .10, .05, .01, and .001. (Round your answers to 4 decimal places.)

	p-value =
	Since p-value =  is less than (Click to select).10, .05, .01.01, .001.001.10 ; reject H0 at those levels of α but not with α = (Click to select).10.001.01, .001.10, .05, .01.

(d) How much evidence is there that the mean composite satisfaction rating exceeds 42?

There is (Click to select)noextremely strongweakvery strongstrong evidence.

Answer 1

(a) Ho: 11 = 42 Ha ll> 42 Z X -li 42.8860. -- 2.5596 on 2.67 1868 12=2.56 Rejection points 2010 = 1.282 Zoos - 1.645 20.01 = 2.326 20.00 - 3.090 Reject to with x = 10.10] (0.05), (0.01) but not with a 0.001) (c) with Z= 2.56 (siglt sailed) we get [P-value = 0.0052 Since Revolve is less than 10.10],[0.05), but nood with a 10.001) (d) is very shoong ] 10.011