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

To determine whether the second sample represents the population of average bowlers, we need to compare the z-score to the critical value at a given level of significance (in this case, p = 0.05).

First, let's calculate the z-score for the second sample mean of 18:

Step 1: Calculate the true standard error of the mean:
σ_(x̅) = σ_X/√N = 6/√30 ≈ 1.0955

Step 2: Transform the sample mean into a z-score using the formula:
Z = (X - µ) / (σ_X/√N) = (18 - 26) / (6/√30) ≈ (-8) / 1.0955 ≈ -7.29

Since the z-score is a very extreme value, it is not available in the z-table. However, we can interpret it as being very far into the tail of the distribution.

Given a significance level of p = 0.05 and using a two-tailed test, we need to find the critical z-value that corresponds to an area of 0.025 in each tail. Looking at the z-table, we find that the critical z-value is approximately ±1.96.

Since the calculated z-score of -7.29 is much more extreme than the critical z-value of ±1.96, we can conclude that the sample does provide sufficient evidence to reject the hypothesis that the second sample represents the population of average bowlers.

In simpler terms, the extreme value of the z-score suggests a very unlikely occurrence, indicating that the second sample is not representative of the population of average bowlers.