91Ó°ÊÓ

British health officials have expressed concern about problems associated with vitamin D deficiency among certain immigrants. Doctors have conjectured that such a deficiency could be related to the amount of fiber in a person's diet. An experiment was designed to compare the vitamin D plasma half-life for two groups of healthy individuals. One group was placed on a normal diet, whereas the second group was placed on a high-fiber diet. The accompanying table gives the resulting data (from "Reduced Plasma Half-Lives of Radio-Labeled \(25(\mathrm{OH}) \mathrm{D} 3\) in Subjects Receiving a High-Fibre Diet," British Journal of Nutrition [1993]: 213-216). \(\begin{array}{lllllll}\text { Normal diet } & 19.1 & 24.0 & 28.6 & 29.7 & 30.0 & 34.8\end{array}\) \(\begin{array}{llllllll}\text { High-fiber diet } & 12.0 & 13.0 & 13.6 & 20.5 & 22.7 & 23.7 & 24.8\end{array}\) Use the following MINITAB output to determine whether the data indicate that the mean half-life is higher for those on a normal diet than those on a high- fiber diet. Assume that treatments were assigned at random and the two plasma half-life distributions are normal. Test the appropriate hypotheses using \(\alpha=.01 .\) Two-sample \(\mathrm{T}\) for normal vs high Iw \(\begin{array}{lccrr} & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } \\ \text { Normal } & 6 & 27.70 & 5.44 & 2.2 \\ \text { High } & 7 & 18.61 & 5.55 & 2.1\end{array}\) \(5.2\) 15 High \(95 \%\) C.l. for mu normal - mu high: \((2.3,15.9)\) T-Test mu normal = mu high(vs >):T \(=2.97 \mathrm{P}=0.0070 \mathrm{DF}=10\)

Step by step solution

01

Formulate Hypotheses

The null hypothesis (\(H_0\)) is that the means of the two populations are equal, that is \(μ_{normal} = μ_{high}\). The alternative hypothesis (\(H_A\)) given the context of the problem is that the mean plasma half-life is higher for the normal diet, so \(μ_{normal} > μ_{high}\)

02

Compare Test Statistic to Critical Value

From the MINITAB output, the test statistic is \(T = 2.97\). The significant level is \(\alpha = 0.01\), so the critical value for a one-tailed t-test with 10 degrees of freedom (obtained using relevant t distribution tables or statistical software) is approximately 2.764. Since our test statistic is greater than the critical value, we reject the null hypothesis.

03

Interpret Results

Because we rejected the null hypothesis, we have evidence to suggest that the mean plasma half-life is higher for people on a normal diet than those on a high-fiber diet. The p-value provided (0.007) is less than our significance level of 0.01, so this finding is statistically significant. Additionally, the 95% confidence interval for the difference in means (2.3, 15.9) does not contain zero, which also supports this finding.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Two-sample t-test

The two-sample t-test is a popular statistical method used to determine whether there is a significant difference between the means of two independent groups. In this case, it helps evaluate if a normal diet results in a higher vitamin D plasma half-life compared to a high-fiber diet.
When performing a two-sample t-test, start by defining your hypotheses. The null hypothesis ( H_0 ) assumes no difference between the means of the two groups, while the alternative hypothesis ( H_A ) suggests that one mean is greater than the other, depending on the context. Here, the alternative hypothesis is that vitamin D half-life is higher for those on a normal diet.
Then, calculate the test statistic. This involves the means of the samples, their standard deviations, and the sample sizes. The test statistic will tell us whether the observed sample mean difference is statistically significant.
The critical value, which you compare with the test statistic, is determined based on the chosen significance level (alpha). If the test statistic exceeds the critical value, you reject the null hypothesis, indicating a significant difference between the groups.

Confidence Interval

A confidence interval provides a range of values that is likely to contain the population parameter, such as the difference in means between two groups. It gives not just a single point estimate but a span of plausible values, making it a valuable tool for understanding statistical significance.
In our example, the confidence interval for the difference in mean vitamin D plasma half-life between a normal and high-fiber diet is (2.3, 15.9). This means we are 95% confident that the true difference in population means falls within this range.
If a confidence interval does not contain zero, it suggests that there is a significant difference between the groups. In this example, since zero is not in the interval, we have strong evidence that the mean half-life is indeed higher for those on a normal diet than for those on a high-fiber diet.
Confidence intervals provide more insight than a simple hypothesis test, as they reflect the variability of the estimate and suggest how reliable this estimate is.

P-value

The p-value is a crucial component of hypothesis testing. It signifies the probability of observing data as extreme as, or more extreme than, the observed sample results, assuming that the null hypothesis is true.
In our dietary study, the calculated p-value is 0.007. This value represents a very low probability under the assumption that there is no difference in the half-life of vitamin D between the two diets, which means the smaller the p-value, the stronger the evidence against the null hypothesis.
Statisticians commonly use a threshold such as 0.01 or 0.05 to determine statistical significance. If the p-value is less than the chosen significance level, we reject the null hypothesis, indicating a statistically significant difference between the groups.
In this analysis, the p-value of 0.007 is less than the significance level of 0.01, leading us to conclude with a high degree of confidence that the vitamin D half-life is indeed higher for individuals on a normal diet compared to those on a high-fiber diet.