Recall that very satisfied customers give the XYZBox video g

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 66 satisfaction ratings to provide evidence supporting the claim that the mean composite satisfaction rating for the XYZ-Box exceeds 42.

Letting µ represent the mean composite satisfaction rating for the XYZ-Box, set up the null hypothesisH₀ and the alternative hypothesis H_a needed if we wish to attempt to provide evidence supporting the claim that µ exceeds 42.

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

  There is: very strong, weak, strong, no, extremely strong evidence

Solution

a)
Set Up Hypothesis
Null H0: U<=42
Alternate H1: U>42
Test Statistic
Population Mean(U)=42
Given That X(Mean)=42.88
Standard Deviation(S.D)=2.69
Number (n)=66
we use Test Statistic (Z) = x-U/(s.d/Sqrt(n))
Zo=42.88-42/(2.69/Sqrt(66)
Zo =2.6577
| Zo | =2.6577
b)
Critical Value
The Value of |Z | at LOS 0.1% is 1.28
The Value of |Z | at LOS 0.05% is 1.64
The Value of |Z | at LOS 0.01% is 2.33
The Value of |Z | at LOS 0.01% is 3.09

Make Decision
Hence Value of | Zo | > | Z | and Here we Reject Ho OTHERWISE Failed to Reject Ho

REJECT HO WHEN ALPHA 0.1,0.05,0.01
BUT N\'T WITH 0.001

c)
P-Value : Right Tail - Ha : ( P > 2.6577 ) = 0.0039
Hence Value of P0.01 > 0.0039, Here we Reject Ho

REJECT HO WHEN ALPHA 0.1,0.05,0.01
BUT N\'T WITH 0.001

d) very strong