Q. 10.47 V. Tangpricha et al. did a study... [FREE SOLUTION]

Chapter 10: Q. 10.47 (page 416)

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/ml ). 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). 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 an increase) in PTH levels, in pg/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 fortified orange juice reduces PTH level more than drinking unfortified orange juice? (Note: The mean and standard deviation for the data on fortified juice are $9.0 pg / mL$ and $37.4 pg / mL$ , respectively, and for the data on unfortified juice, they are $1.6 pg / mL$ and $34.6 pg / mL$ , respectively.)

Short Answer

Expert verified

The supplied data do not provide adequate evidence to indicate that drinking fortified orange juice lowers PTH levels more than drinking unfortified orange juice at the significance level of $5 %$ .

Step by step solution

Given Information

Sample data for the decrease in PTH level from two populations of fortified and unfortified orange juice is presented.

${\overset{炉}{x}}_{1} = 9.0, s_{1} = 37.4$

${\overset{炉}{x}}_{2} = 1.6, s_{2} = 34.6$

The significance level is $5 %$ ,

Explanation

Population $1$ : Fortified orange juice, ${\overset{炉}{x}}_{1} = 9.0, s_{1} = 37.4, and n_{1} = 14$ .

Population $2$ : Unfortified orange juice, ${\overset{炉}{x}}_{2} = 1.6, s_{2} = 34.6, and n_{2} = 12$ .

The main goal is to determine whether drinking fortified orange juice lowers PTH levels more than drinking unfortified orange juice on average. This means that after drinking fortified orange juice, the average fall in PTH level should be greater than after drinking unfortified orange juice.

Define null and alternate hypotheses.

Null hypotheses: $H_{0} = 渭_{1} 鈮� 渭_{2}$

Alternate hypotheses: $H_{a} = 渭_{1} > 渭_{2}$

Hypotheses is right-tailed.

We determine significance level as $5 %$ .

Calculation

Pooled standard deviation, $s_{p} = \sqrt{\frac{(n_{1} - 1) s_{1}^{2} + (n_{2} - 1) s_{2}^{2}}{n_{1} + n_{2} - 2}}$

$鈬� s_{p} = \sqrt{\frac{(14 - 1) (37.4)^{2} + (12 - 1) (34.6)^{2}}{14 + 12 - 2}}$

$鈬� s_{p} = \sqrt{\frac{13 (1398.76) + 11 (1197.16)}{24}}$

$鈬� s_{p} = 36.144$

Test statistic, $t_{0} = \frac{{\overset{炉}{x}}_{1} - {\overset{炉}{x}}_{2}}{s_{p} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}}$

$鈬� t_{0} = \frac{9.0 - 1.6}{36.144 \sqrt{\frac{1}{14} + \frac{1}{12}}}$

$鈬� t_{0} = 0.520$

We need to decide the critical values

Here, localid="1651301477365" role="math" $d f = n_{1} + n_{2} - 2 = 14 + 12 - 2 = 24$

localid="1651301491754" $鈬� d f = 24$

Using table IV

localid="1651301486241" role="math" $Critical value, t_{伪} = t_{0.05} = 1.711$

Since the test statistic, $t_{0} < t_{伪}$ , does not fall in the rejection zone of the left-tailed hypotheses test. As a result, null hypotheses are not ruled out.