Question

Preliminary data analyses indicate that you can reasonably consider the assumptions for using pooled $t$ -procedures satisfied. For each exercise, perform the required hypothesis test by using either the critical-value approach or the P-value approach. Fortified Juice and PTH. V. Tangpricha et al. did a study to determine whether fortifying orange juice with Vitamin D would result in changes in the blood levels of five biochemical variables. One of those variables was the concentration of parathyroid hormone (PTH), measured in picograms/milliliter (pg/mil). The researchers published their results in the paper "Fortification of Orange Juice with Vitamin D: A Novel Approach for Enhancing Vitamin D Nutritional Health" (American Journal of Clinical Nutrition, Vol. 77, pp. 1478-1483). A doubleblind experiment was used in which 14 subjects drank $240 \mathrm{mL}$ per day of orange juice fortified with 1000 IU of Vitamin D and 12 subjects drank 240 mL. per day of unfortified orange juice. Concentration levels were recorded at the beginning of the experiment and again at the end of 12 weeks. The following data, based on the results of the study, provide the decrease (negative values indicate increase) in PTH levels, in $\mathrm{pg} / \mathrm{mL}$, for those drinking the fortified juice and for those drinking the unfortified juice. At the $5 \%$ significance level, do the data provide sufficient evidence to conclude that drinking fortificd orange juice reduces PTH level more than drinking unfortified orange juice? (Note: The mean and standard deviation for the data on fortified juice are $9.0 \mathrm{pg} / \mathrm{mL}$ and $37.4 \mathrm{pg} / \mathrm{mL},$ respectively, and for the data on unfortified juice, they are $1.6 \mathrm{pg} / \mathrm{mL}$ and $34.6 \mathrm{pg} / \mathrm{mL}$, respectively.)

Name: Problem 41, Chapter 10: Introductory Statistics
Uploaded: 2021-07-02T16:30:36-08:00
Duration: 3 min 7 s
Channel: Maxime Rossetti

   Preliminary data analyses indicate that you can reasonably consider the assumptions for using pooled $t$ -procedures satisfied. For each exercise, perform the required hypothesis test by using either the critical-value approach or the P-value approach.
Fortified Juice and PTH. V. Tangpricha et al. did a study to determine whether fortifying orange juice with Vitamin D would result in changes in the blood levels of five biochemical variables. One of those variables was the concentration of parathyroid hormone (PTH), measured in picograms/milliliter (pg/mil). The researchers published their results in the paper "Fortification of Orange Juice with Vitamin D: A Novel Approach for Enhancing Vitamin D Nutritional Health" (American Journal of Clinical Nutrition, Vol. 77, pp. 1478-1483). A doubleblind experiment was used in which 14 subjects drank $240 \mathrm{mL}$ per day of orange juice fortified with 1000 IU of Vitamin D and 12 subjects drank 240 mL. per day of unfortified orange juice. Concentration levels were recorded at the beginning of the experiment and again at the end of 12 weeks. The following data, based on the results of the study, provide the decrease (negative values indicate increase) in PTH levels, in $\mathrm{pg} / \mathrm{mL}$, for those drinking the fortified juice and for those drinking the unfortified juice. At the $5 \%$ significance level, do the data provide sufficient evidence to conclude that drinking fortificd orange juice reduces PTH level more than drinking unfortified orange juice? (Note:
The mean and standard deviation for the data on fortified juice are $9.0 \mathrm{pg} / \mathrm{mL}$ and $37.4 \mathrm{pg} / \mathrm{mL},$ respectively, and for the data on unfortified juice, they are $1.6 \mathrm{pg} / \mathrm{mL}$ and $34.6 \mathrm{pg} / \mathrm{mL}$, respectively.)

Introductory Statistics

Neil A. Weiss 9th Edition

Chapter 10, Problem 41

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Solved by Expert Maxime Rossetti

07/02/2021

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The null hypothesis is that there is no difference in the decrease in PTH levels between those drinking fortified orange juice and those drinking unfortified orange juice. The alternative hypothesis is that drinking fortified orange juice reduces PTH level more Show more…

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Preliminary data analyses indicate that you can reasonably consider the assumptions for using pooled $t$ -procedures satisfied. For each exercise, perform the required hypothesis test by using either the critical-value approach or the P-value approach. Fortified Juice and PTH. V. Tangpricha et al. did a study to determine whether fortifying orange juice with Vitamin D would result in changes in the blood levels of five biochemical variables. One of those variables was the concentration of parathyroid hormone (PTH), measured in picograms/milliliter (pg/mil). The researchers published their results in the paper "Fortification of Orange Juice with Vitamin D: A Novel Approach for Enhancing Vitamin D Nutritional Health" (American Journal of Clinical Nutrition, Vol. 77, pp. 1478-1483). A doubleblind experiment was used in which 14 subjects drank $240 \mathrm{mL}$ per day of orange juice fortified with 1000 IU of Vitamin D and 12 subjects drank 240 mL. per day of unfortified orange juice. Concentration levels were recorded at the beginning of the experiment and again at the end of 12 weeks. The following data, based on the results of the study, provide the decrease (negative values indicate increase) in PTH levels, in $\mathrm{pg} / \mathrm{mL}$, for those drinking the fortified juice and for those drinking the unfortified juice. At the $5 \%$ significance level, do the data provide sufficient evidence to conclude that drinking fortificd orange juice reduces PTH level more than drinking unfortified orange juice? (Note: The mean and standard deviation for the data on fortified juice are $9.0 \mathrm{pg} / \mathrm{mL}$ and $37.4 \mathrm{pg} / \mathrm{mL},$ respectively, and for the data on unfortified juice, they are $1.6 \mathrm{pg} / \mathrm{mL}$ and $34.6 \mathrm{pg} / \mathrm{mL}$, respectively.)

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Key Concepts

Hypothesis Testing

Hypothesis testing is a statistical method used to decide whether there is enough evidence in a sample of data to infer that a certain condition holds for the entire population. It typically involves formulating a null hypothesis (that there is no effect or difference) and an alternative hypothesis (that there is an effect or difference) and then using data to test these hypotheses.

Null and Alternative Hypotheses

The null hypothesis is a default statement asserting that there is no change or effect, while the alternative hypothesis represents the outcome that the researcher suspects might be true. Establishing these hypotheses is essential for structuring the testing procedure and determining the direction of the test.

Two-Sample t-Test

A two-sample t-test is used to compare the means of two independent groups to determine if there is a statistically significant difference between them. This test is particularly useful when the population variances are assumed to be equal and the sample sizes are relatively small.

Pooled Variance

Pooled variance is a method for estimating the common variance of two populations when it is assumed that both populations have the same variance. This estimate is calculated by combining the variances from both samples, which provides a more accurate measure for the t-test statistic in cases of equal variance.

t-Distribution

The t-distribution is a probability distribution that is used in hypothesis testing when the sample size is small and the population standard deviation is unknown. It resembles the standard normal distribution but has heavier tails, meaning it accounts for the increased uncertainty in smaller samples.

Degrees of Freedom

Degrees of freedom refer to the number of values in a calculation that are free to vary when computing a statistic. In the context of a pooled two-sample t-test, degrees of freedom are typically calculated as the total sample size minus two, which influences the shape of the t-distribution used for determining critical values.

Critical Value Approach

The critical value approach involves comparing the test statistic to a threshold value, derived from the appropriate probability distribution, at a specified significance level. If the test statistic falls in the critical region, the null hypothesis is rejected.

P-Value Approach

The p-value approach calculates the probability of obtaining a test statistic as extreme as, or more extreme than, the observed value under the assumption that the null hypothesis is true. This p-value is then compared to a predetermined significance level to decide whether to reject the null hypothesis.

Significance Level

The significance level, often set at 5% (0.05), is the probability threshold used to determine whether an observed effect is statistically significant. If the p-value is less than the significance level, researchers reject the null hypothesis, indicating that the observed effect is unlikely to have occurred by chance alone.

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