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 65 satisfaction ratings to provide evidence supporting he 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 H₀ needed if we wish to attempt to provide evidence supporting the claim that µ exceeds 42.
H₀: µ ≤ 42
H₀: µ > 42
The random sample of 65 satisfaction ratings yields a sample mean of Ẋ = 42.954. Assuming that σ equals 2.64 use critical values to test H₀ versus H₀ at each of α = .10, .05, .001, and .001.
z = (Ẋ - µ ) / (σ/√n)
z = (42.954 – 42) /(2.64 / √65)
z= .954 / (2.64/ 8.0623)
z = 2.9134
α = z critical result
Using the information in part b, calculate the p-value and use it to text H₀ versus H₀ at each of α = .10, .05, .01 and .001.
p – value is .0025
The result is significant at .10, .05, and .01, because p < α at .001, the result is not significant because p > α d.
How much evidence is there that the mean composite satisfaction rating exceeds 42? We can be at least 99% certain, but not as much as 99.9 certain.
How do we decide whether to use a z test or a t test when testing a hypothesis about a population mean? Assuming the population is approximately normally distributed, we can use the z-test when the sample size is large or knowing the population standard deviation σ, even if the population size is small (n ≥ 30). However if the sample size is small and all the sample standard deviation then we must use the t – distribution.
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