The Following Information Is For Questions 1 Through 6the Recommended
The following information is for Questions 1 through 6. The recommended daily allowance (RDA) of cobalamine (Vitamin B12) for growing teens is 2.4 µg. It is generally believed that growing teens are getting less than the RDA of 2.4 µg of cobalamine daily. Médecins Sans Frontières (MSF) volunteers collected blood samples from 10 randomly selected teens in a developing country, measuring their cobalamine levels (in µg) as follows: 1.85, 2.35, 1.87, 1.90, 1.37, 2.35, 2.55, 2.28, 1.95, 2.49. The population standard deviation of cobalamine in teens is assumed to be 0.56 µg based on MSF’s global experience. Determine if these teens are getting the recommended daily allowance of cobalamine, considering possible dispute from the host country.
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To evaluate whether the teens are receiving adequate cobalamine intake, a hypothesis test must be conducted. Given that the population standard deviation is known and the sample size is small (n=10), a z-test is appropriate for this analysis. The z-test is used when the population variance is known, and the sample size is sufficiently small or large; here, since the standard deviation is given, it justifies the use of the z-test.
The null hypothesis (H■) claims that the mean cobalamine level in teens equals the RDA of 2.4 µg, while the alternative hypothesis (H■) asserts that the mean level is less than 2.4 µg, indicating insufficient intake. This results in a left-tailed test because we are testing whether the mean is less than the RDA:
H■: µ = 2.4
H■: µ < 2.4
To compute the test statistic, we calculate the sample mean (\(\bar{x}\)) from the data:
\(\bar{x} = \frac{1.85 + 2.35 + 1.87 + 1.90 + 1.37 + 2.35 + 2.55 + 2.28 + 1.95 + 2.49}{10} = 2.008\) µg
The standard error (SE) for the z-test is:
SE = \(\frac{\sigma}{\sqrt{n}} = \frac{0.56}{\sqrt{10}} ≈ 0.177\)
The z-statistic is then calculated as:
z = \(\frac{\bar{x} - \mu_0}{SE} = \frac{2.008 - 2.4}{0.177} ≈ -2.17\)
Using standard normal distribution tables or statistical software, the p-value associated with z = -2.17 is approximately 0.015.
Since the p-value (≈0.015) is less than the common significance level of 0.05, we reject the null hypothesis. This suggests that the average cobalamine level in the sample of teens is statistically significantly less than the RDA, indicating inadequate intake.
If I were the representative of the host country, I would argue that individual variation and measurement errors could account for the lower observed levels. I would emphasize that the sample size is small and that broader testing might be necessary to make definitive conclusions. If I support MSF's findings, I would state that the statistical evidence indicates a real deficiency, which warrants policy intervention to improve nutrition. Conversely, if I support the host country, I might argue that the differences are within expected variability, and further, more extensive sampling is needed before drawing firm conclusions.
If the population standard deviation were unknown, an alternative approach would be to use a t-test. The t-test would replace the z-statistic with:
t = \(\frac{\bar{x} - \mu_0}{s / \sqrt{n}}\)
where s is the sample standard deviation calculated from the data. Performing this t-test involves computing s, degrees of freedom (df = n-1), and referencing the t-distribution to find the p-value, offering a more conservative test when the population standard deviation isn't available.
Question 7: Testing Difference Between Two Groups
The second group of data from the host country shows cobalamine levels: 2.45, 2.85, 2.87, 2.32, 1.98, 2.51, 1.75, 1.98, 2.03, 2.89. To test whether these two groups' means are statistically different, assuming unknown population variances, a two-sample t-test is appropriate.
The null hypothesis (H■): µ■ = µ■, indicating no difference in mean cobalamine levels between the MSF and host country groups. The alternative hypothesis (H■): µ■ ≠ µ■, suggesting a significant difference in means. Both sides are considered (two-tailed test). The test involves calculating the sample means and variances, then computing the t-statistic and associated p-value accordingly.
Question 8: Probability of Type I error
At a significance level (α) of 0.05, the probability of committing a Type I error—that is, rejecting the null
hypothesis when it is actually true—is exactly 5%. This is the predefined alpha level set by the researcher, reflecting the acceptable risk of false positives.
Question 9: Power and Type II Error for Different RDA Values
If the true RDA varies across a range from 2.0 to 2.6 µg, the statistical test's power (the probability of correctly rejecting H■ when it is false) depends on the true mean. For each assumed RDA, we can calculate the non-centrality parameter and then the power and Type II error (β). These calculations involve the effect size (difference from H■), the standard error, and the significance level. Generally, larger differences from the null hypothesis increase the statistical power.
Question 10: Plotting Power Curve and Observations
A power curve plots the probability of correctly rejecting H■ (power) against the hypothesized true means across various RDA levels. As the true mean moves further away from the null value (2.4 µg), the power increases, indicating higher likelihood of detecting a real difference. The observed pattern confirms that the test is more sensitive to larger deviations, emphasizing the importance of the chosen significance level in balancing Type I and Type II errors.
Questions 11 & 12: Confidence Intervals and Testing Support for Climate Change
In the poll of 500 adults, 420 responded “Yes,” leading to a sample proportion (\(\hat{p}\)) of 0.84. A 90% confidence interval for the true proportion is calculated using the critical z-value of approximately 1.645. The interval is:
\(\hat{p} \pm z_{0.05} \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = 0.84 \pm 1.645 \times \sqrt{\frac{0.84 \times 0.16}{500}}\)
Calculating the margin of error yields approximately 0.028, resulting in a confidence interval of (0.812, 0.868). This interval estimates that between 81.2% and 86.8% of adults support the claim with 90% confidence.
To assess whether the proportion of supporters has increased from 2016 (80%) to 2017 (84%), a two-proportion z-test is performed. The test statistic is:
z = \(\frac{\hat{p}_1 - \hat{p}_2}{\sqrt{p(1-p)(\frac{1}{n_1} + \frac{1}{n_2})}}\), where p is the pooled proportion.
Calculating this yields a z-value of approximately 1.66 and an associated p-value around 0.096. Since this p-value exceeds α=0.05, there is insufficient evidence to conclude that support has significantly increased.
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