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Chapter 11: Problem 17

Low-birth-weight babies are at increased risk of respiratory infections in the first few months of life and have low liver stores of vitamin A. In a randomized, double-blind experiment, 130 lowbirth-weight babies were randomly divided into two groups. Subjects in group 1 (the treatment group, \(n_{1}=65\) ) were given 25,000 IU of vitamin A on study days \(1,4,\) and 8 where study day 1 was between 36 and 60 hours after delivery. Subjects in group 2 (the control group, \(n_{2}=65\) ) were given a placebo. The treatment group had a mean serum retinol concentration of 45.77 micrograms per deciliter \((\mu \mathrm{g} / \mathrm{dL}),\) with a standard deviation of \(17.07 \mu \mathrm{g} / \mathrm{dL}\). The control group had a mean serum retinol concentration of \(12.88 \mu \mathrm{g} / \mathrm{dL}\), with a standard deviation of \(6.48 \mu \mathrm{g} / \mathrm{dL}\). Does the treatment group have a higher standard deviation for serum retinol concentration than the control group at the \(\alpha=0.01\) level of significance? It is known that serum retinol concentration is normally distributed.

Short Answer

Expert verified
The treatment group has a significantly higher standard deviation for serum retinol concentration at the 伪 = 0.01 level of significance.

Step by step solution

01

State the null and alternative hypotheses

Formulate the hypotheses for comparing the variances. The null hypothesis (H_0) and the alternative hypothesis (H_A) are: H_0: (蟽^2_1 = 蟽^2_2) where 蟽^2_1 and 蟽^2_2 are the population variances of the treatment group and control group, respectively. H_A: (蟽^2_1 > 蟽^2_2). This is a one-tailed test.
02

Calculate the F-statistic

Find the ratio of the sample variances to determine the F-statistic. Calculate the sample variances (s^2_1 and s^2_2). s^2_1 = (17.07)^2 and s^2_2 = (6.48)^2. Therefore, s^2_1 = 291.8569 (渭g / dL)^2 and s^2_2 = 41.9904 (渭g / dL)^2. The F-statistic is then F = (s^2_1 / (s^2_2)). Thus, F = (291.8569 / 41.9904) 鈮� 6.95.
03

Determine the critical value from the F-distribution table

Using the F-distribution table, find the critical value for a one-tailed test at 伪 = 0.01 with degrees of freedom (df_1 = 64) and (df_2 = 64). Look up the critical value in the F-table, which corresponds to F_critical = 2.21.
04

Compare the F-statistic to the critical value

Compare the calculated F-statistic to the critical value to make a decision. Since F = 6.95 is greater than F_critical = 2.21, reject the null hypothesis (H_0).
05

Draw a conclusion

Based on the results of the comparison, there is sufficient evidence to conclude that the treatment group has a higher standard deviation for serum retinol concentration compared to the control group at the 伪 = 0.01 level of significance.

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Key Concepts

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

randomized double-blind experiment
A randomized double-blind experiment is a type of scientific study designed to reduce bias and increase reliability. In this type of experiment, participants are randomly assigned to either the treatment or control group. This helps ensure that the groups are as similar as possible so that any differences observed are likely due to the treatment itself. Additionally, neither the participants nor the experimenters know who is receiving the treatment and who is receiving a placebo. This 'double-blind' setup prevents bias from affecting the results, as neither side can influence the outcomes based on prior knowledge or expectations. In our exercise, low-birth-weight babies were divided into two groups: one receiving vitamin A and the other a placebo. This setup aimed to objectively assess the impact of vitamin A on serum retinol concentration.
F-statistic
The F-statistic is a crucial component in hypothesis testing, especially when comparing variances between two groups. It is the ratio of two sample variances and is used to determine if the observed differences between groups are statistically significant. The formula for the F-statistic is \(F = \frac{s^2_1}{s^2_2}\), where \(s^2_1\) and \(s^2_2\) are the sample variances of the two groups. In our exercise, the variances were calculated as \(291.8569 (\mu g / dL)^2\) for the treatment group and \(41.9904 (\mu g / dL)^2\) for the control group. This resulted in an F-statistic of \approx 6.95\. A higher F-statistic indicates a greater degree of variation between the sample groups' variances, suggesting a potential difference in the populations they represent.
critical value
The critical value in hypothesis testing is a threshold that determines whether to accept or reject the null hypothesis. It is derived from the F-distribution table based on the desired level of significance (\(\alpha\)) and the degrees of freedom (df) for both the numerator and the denominator. For our exercise, with \(\alpha = 0.01\), and degrees of freedom \(df_1 = 64\) and \(df_2 = 64\), the critical value was found to be \approx 2.21\. This means that if our calculated F-statistic exceeds 2.21, we can reject the null hypothesis, suggesting that there is statistically significant evidence that the variances are different. Since our F-statistic of 6.95 is much higher than this critical value, we reject the null hypothesis.
serum retinol concentration
Serum retinol concentration refers to the amount of vitamin A present in the blood. It is a critical measure for assessing vitamin A status and is particularly important for vulnerable populations, such as low-birth-weight infants who are at risk of vitamin A deficiency. In our study, the treatment group received vitamin A, and we measured their serum retinol concentration to assess the effectiveness of the supplementation. The mean serum retinol concentration was found to be \(45.77 \mu g / dL\) for the treatment group and \(12.88 \mu g / dL\) for the control group. By comparing these measurements, researchers can determine whether vitamin A supplementation effectively increases serum retinol levels in the treatment group.

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Most popular questions from this chapter

Stock fund managers are investment professionals who decide which stocks should be part of a portfolio. In a recent article in the Wall Street Journal ( \(N\) ot \(a\) Stock-Picker's Market, WSJ January 25,2014 ), the performance of stock fund managers was considered based on dispersion in the market. In the stock market, risk is measured by the standard deviation rate of return of stock (dispersion). When dispersion is low, then the rate of return of the stocks that make up the market are not as spread out. That is, the return on Company \(X\) is close to that of \(Y\) is close to that of \(Z,\) and so on. When dispersion is high, then the rate of return of stocks is more spread out; meaning some stocks outperform others by a substantial amount. Since \(1991,\) the dispersion of stocks has been about \(7.1 \% .\) In some years, the dispersion is higher (such as 2001 when dispersion was \(10 \%)\), and in some years it is lower (such as 2013 when dispersion was \(5 \%)\). So, in 2001 , stock fund managers would argue, one needed to have more investment advice in order to identify the stock market winners, whereas in 2013 , since dispersion was low, virtually all stocks ended up with returns near the mean, so investment advice was not as valuable. (a) Suppose you want to design a study to determine whether the proportion of fund managers who outperform the market in low-dispersion years is less than the proportion of fund managers who outperform the market in high-dispersion years. What would be the response variable in this study? What is the explanatory variable in this study? (b) What or who are the individuals in this study? (c) To what population does this study apply? (d) What would be the null and alternative hypothesis? (e) Suppose this study was conducted and the data yielded a \(P\) -value of \(0.083 .\) Explain what this result suggests.

In an experiment conducted online at the University of Mississippi, study participants are asked to react to a stimulus. In one experiment, the participant must press a key on seeing a blue screen and reaction time (in seconds) to press the key is measured. The same person is then asked to press a key on seeing a red screen, again with reaction time measured. The results for six randomly sampled study participants are as follows: $$ \begin{array}{lcccccc} \text { Participant } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} \\ \hline \text { Blue } & 0.582 & 0.481 & 0.841 & 0.267 & 0.685 & 0.450 \\ \hline \text { Red } & 0.408 & 0.407 & 0.542 & 0.402 & 0.456 & 0.533 \\ \hline \end{array} $$ (a) Why are these matched-pairs data? (b) In this study, the color that the participant was first asked to react to was randomly selected. Why is this a good idea in this experiment? (c) Is the reaction time to the blue stimulus different from the reaction time to the red stimulus at the \(\alpha=0.01\) level of significance? Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers. (d) Construct a \(99 \%\) confidence interval about the population mean difference. Interpret your results. (e) Draw a boxplot of the differenced data. Does this visual evidence support the results obtained in part (c)?

The Harris Poll conducted a survey in which they asked, 鈥淗ow many tattoos do you currently have on your body?鈥� Of the 1205 males surveyed, 181 responded that they had at least one tattoo. Of the 1097 females surveyed, 143 responded that they had at least one tattoo. Construct a 95% confidence interval to judge whether the proportion of males that have at least one tattoo differs significantly from the proportion of females that have at least one tattoo. Interpret the interval.

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