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 64 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 64 satisfaction ratings yields a sample mean of x¯=42.970x¯=42.970. 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).001.10.10, .05, .01.01, .001 , but not with α =(Click to select).10.10, .05, .01.01, .001.001

(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)very strongextremely strongnoweakstrong evidence.

Answer 1

a) H₀: < 42

H_a: > 42

b) At = 0.10, the critical value is z_0.9 = 1.28

At = 0.05, the critical value is z_0.95 = 1.645

At = 0.01, the critical value is z_0.99 = 2.33

At = 0.001, the critical value is z_0.999 = 3.08

The test statistic z = ()/()

= (42.97 - 42)/(2.67/)

= 2.91

Reject H₀, if Z > z_critcritical

Reject H₀ with = 0.10, 0.05, 0.01, but not with = 0.001

c) P-value = P(Z > 2.91)

= 1 - P(Z < 2.91)

= 1 - 0.9982

= 0.0018

Reject H₀, if P-value is less than .

Since the P-value = 0.0018 is less than 0.10, 0.05, 0.01; reject H₀ at those levels of but not with = 0.001

There is very strong evidence that the mean composite satisfaction rating exceeds 42 at = 0.10, 0.05, 0.01.

There is weak evidence that the mean composite satisfaction rating exceeds 42 at = 0.001.