Step 2 - Transform the sample mean into a z-score (p. 133)
x̅ = 26, N = 30, σ_X=6
Z = (X- µ)/(σ_X/√N) = (24- 26)/(6/5.4772) = (-2)/(1.0955) = 1.8257
Z = - 1.83
The z-score table notes that the area between the mean of 26 and the z-score of 1.83 = .4664. The area beyond z in the tail = .0336. Since .0336 is greater than our criterion of 0.025, we do not have sufficient evidence that concludes the sample does not represent the population of bowlers.
2. A second random sample of 30 bowlers at Ethel’s Bowling Alley has a mean of 18. x̅ = 18, N = 30, σ_X=6
Step 1 - Calculate the true standard error of the mean.
σ_(x̅) =σ_X/√N = 18/√6 = 18/2.45 = 7.35 , or 7
σ_(x̅) =7.35 ,or 7
Step 2 - Transform the sample mean into a z-score (p. 133)
Z = (X- µ)/(σ_X/√N) = (24- 18)/(6/5.4772) = (+ 6)/(1.0955) = 5.48
Z = 5.48
The z-score table notes that the area between the mean of 18 and the z-score of 5.48 = .4664. The area beyond z in the tail = .0336. Since .0336 is less than our criterion of 0.025, we do not have sufficient evidence that concludes the sample does not represent the population of bowlers.
Using the criterion of p = .05 and both tails of the sampling distribution, decide if each sample represents the population of the average bowlers.
Using the z-table, we find that .025 of the curve lies beyond the z-score of -1.96
I can’t find a z-score of 5.48 on the z-table. Does this sample represent the population?
Much thanks,
Mari
1 answer