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:
For Scenario 1, where we are given the population mean (29.6) and standard deviation (7.5), we will use a Z-value because the population standard deviation is known. The Z-value formula is: Z=X‾−μσnZ = \frac{\overline{X} – \mu}{\frac{\sigma}{\sqrt{n}}}Z=nσX−μ
Where:
- X‾\overline{X}X = sample mean
- μ\muμ = population mean
- σ\sigmaσ = population standard deviation
- nnn = sample size
Sample 1:
- Sample size (nnn) = 30
- Sample mean (X‾\overline{X}X) = 31.8
- Population mean (μ\muμ) = 29.6
- Population standard deviation (σ\sigmaσ) = 7.5
Substitute into the Z formula: Z=31.8−29.67.530=2.27.55.477=2.21.37=1.61Z = \frac{31.8 – 29.6}{\frac{7.5}{\sqrt{30}}} = \frac{2.2}{\frac{7.5}{5.477}} = \frac{2.2}{1.37} = 1.61Z=307.531.8−29.6=5.4777.52.2=1.372.2=1.61
Sample 2:
- Sample size (nnn) = 50
- Sample mean (X‾\overline{X}X) = 31.6
- Population mean (μ\muμ) = 29.6
- Population standard deviation (σ\sigmaσ) = 7.5
Substitute into the Z formula: Z=31.6−29.67.550=27.57.071=21.06=1.89Z = \frac{31.6 – 29.6}{\frac{7.5}{\sqrt{50}}} = \frac{2}{\frac{7.5}{7.071}} = \frac{2}{1.06} = 1.89Z=507.531.6−29.6=7.0717.52=1.062=1.89
Are these samples concerning?
The Z-values of 1.61 and 1.89 suggest that the samples are moderately higher than the population mean, but they are not extreme. In a normal distribution, a Z-value of 1.96 or greater would typically be considered significantly different from the population mean. Since both Z-values are below 1.96, neither sample is overly concerning, but we should still monitor any patterns of consistently higher BMIs.
Scenario 2:
For Scenario 2, since the population standard deviation is not provided and we are using sample data, we will use a T-value instead of the Z-value. The formula for the T-value is: T=X‾−μsnT = \frac{\overline{X} – \mu}{\frac{s}{\sqrt{n}}}T=nsX−μ
Where:
- X‾\overline{X}X = sample mean
- μ\muμ = population mean
- sss = sample standard deviation
- nnn = sample size
Sample 1:
- Sample size (nnn) = 30
- Sample mean (X‾\overline{X}X) = 27.6
- Population mean (μ\muμ) = 29.6
- Sample standard deviation (sss) = 6.8
Substitute into the T formula: T=27.6−29.66.830=−26.85.477=−21.24=−1.61T = \frac{27.6 – 29.6}{\frac{6.8}{\sqrt{30}}} = \frac{-2}{\frac{6.8}{5.477}} = \frac{-2}{1.24} = -1.61T=306.827.6−29.6=5.4776.8−2=1.24−2=−1.61
Sample 2:
- Sample size (nnn) = 50
- Sample mean (X‾\overline{X}X) = 28.0
- Population mean (μ\muμ) = 29.6
- Sample standard deviation (sss) = 7.7
Substitute into the T formula: T=28.0−29.67.750=−1.67.77.071=−1.61.09=−1.47T = \frac{28.0 – 29.6}{\frac{7.7}{\sqrt{50}}} = \frac{-1.6}{\frac{7.7}{7.071}} = \frac{-1.6}{1.09} = -1.47T=507.728.0−29.6=7.0717.7−1.6=1.09−1.6=−1.47
Are these samples concerning?
The T-values for both samples (-1.61 and -1.47) suggest that the sample means are lower than the population mean, but they are not drastically different. Like with the Z-values, a T-value below -2 or above 2 would indicate a more significant difference. Therefore, these results are not highly concerning but could indicate some variability in BMI among the patients suffering from gout. It could be worth investigating whether dietary, genetic, or environmental factors are contributing to these variations.
