Instructions For this discussion post, we are going to practice finding Z and T values for sets of data. We have two scenarios to look at: Scenario 1 – The average BMI of patients suffering from gout is known to follow a Normal distribution with a mean of 29.6 with a standard deviation of 7.5. We decide to sample some people suffering from gout and find the following from our samples: Sample 1 – Sample size of 30 with a Mean BMI = 31.8 Sample 2 – Sample size of 50 with a Mean BMI = 31.6 Scenario 2 – The average BMI of patients suffering from gout is known to follow a Normal distribution with a mean of 29.6, but no standard deviation was reported. We decide to sample some people suffering from gout and find the following from our samples: Sample 1 – Sample size of 30 with a Mean BMI = 27.6, and a sample standard deviation of 6.8 Sample 2 – Sample size of 50 with a Mean BMI = 28.0, and a sample standard deviation of 7.7 Discussion Prompts Answer the following questions in your initial post: For Scenario 1, will you choose to compute a Z or a T value? Explain your reasoning Compute the Z or T value for each one of your samples. Are either of these samples concerning to us? If so, what reasons may attribute to something like this? For Scenario 2, will you choose to compute a Z or a T value? Explain your reasoning Compute the Z or T value for each one of your samples. Are either of these samples concerning to us? If so, what reasons may attribute to something like this?
The Correct Answer and Explanation is:
Scenario 1 – Known Population Standard Deviation
Since the population standard deviation is known (7.5), we use the Z-score for this scenario. Z-scores are appropriate when the population standard deviation is known, regardless of the sample size.
Sample 1 – n = 30, Mean = 31.8
Z = (X̄ – μ) / (σ / √n)
Z = (31.8 – 29.6) / (7.5 / √30)
Z = 2.2 / (7.5 / 5.477)
Z = 2.2 / 1.369
Z ≈ 1.61
Sample 2 – n = 50, Mean = 31.6
Z = (31.6 – 29.6) / (7.5 / √50)
Z = 2.0 / (7.5 / 7.071)
Z = 2.0 / 1.061
Z ≈ 1.88
Are they concerning?
Both Z-scores (1.61 and 1.88) are below 2, which means they are not statistically significant at the 0.05 level. This suggests that while the mean BMIs are slightly higher than the population mean, they are within an acceptable range and not concerning.
Scenario 2 – Unknown Population Standard Deviation
Since the population standard deviation is unknown and we are using sample data, we use the T-score. T-scores are used when the population standard deviation is not known and the sample standard deviation is used instead.
Sample 1 – n = 30, Mean = 27.6, s = 6.8
T = (X̄ – μ) / (s / √n)
T = (27.6 – 29.6) / (6.8 / √30)
T = -2.0 / (6.8 / 5.477)
T = -2.0 / 1.241
T ≈ -1.61
Sample 2 – n = 50, Mean = 28.0, s = 7.7
T = (28.0 – 29.6) / (7.7 / √50)
T = -1.6 / (7.7 / 7.071)
T = -1.6 / 1.089
T ≈ -1.47
Are they concerning?
T-scores of -1.61 and -1.47 do not exceed the critical value of approximately ±2 for 95 percent confidence. So these are not statistically significant and not alarming. Slight deviations may occur due to random sampling or demographic differences.
