Question

Fortified Juice and PIH. Refer to Exercise 10.47 and find a 90% confidence interval for the difference between the mean reductions in PTH levels for fortified and unfortified orange juice.

Short answer

We can be $90\%$positive that fortified orange juice reduces PTH levels by$16.92pg/ml$or$31.72pg/ml$compared to unfortified orange juice.

Step-by-step solution

Given Information

Given in the question that,

PTH level decrease sample data from two populations of fortified orange juice and unfortified orange juice are provided.

${\overline{x}}_1=9.0$

$s_1=37.4$

${\overline{x}}_2=1.6$

$s_2=34.6$

The significance level is$5\%$.

Explanation

Let's consider the Population 1:

Fortified orange juice,${\overline{x}}_1=9.0$

$s_1=37.4$and

$n_1=14$

Let's consider the Population 2:

Unfortified orange juice,${\overline{x}}_2=1.6$

$s_2=34.6$and

$n_2=12$

The main goal is to determine a $90\%$confidence interval for the difference between two population means, ${\mu }_1$and ${\mu }_2$

The null and alternate hypotheses should be stated.

Null hypotheses: $H_0={\mu }_1\le {\mu }_2$

Alternate hypotheses: $H_a={\mu }_1>{\mu }_2$

Hypotheses are right-tailed.

Table IV may be used to find $t_{\alpha /2}$with $df=n_1+n_2-2$with a confidence level of $1-a$.localid="1651401100098" role="math" $a=0.10$for 90% confidence level

localid="1651401120491" $$\begin{gathered} \text{df}=n_1+n_2-2 \\ =(14+12-2) \\ =24. \end{gathered}$$

When $\mathrm{df}=24$using table IV for critical values,

Critical value,

localid="1651401168912" $$\begin{gathered} t_{\alpha /2}=t_{0.10/2} \\ =t_{0.05} \\ =1.711 \end{gathered}$$

The confidence interval's endpoints must be found.

Pooled standard deviation,

$s_p=\sqrt{\frac{\left(n_1-1\right)s_1{}^2+\left(n_2-1\right){s_2}^2}{n_1+n_2-2}}$

$s_p=\sqrt{\frac{(14-1)(37.4{)}^2+(12-1\left)\right(34.6{)}^2}{14+12-2}}$

$s_p=\sqrt{\frac{13(1398.76)+11(1197.16)}{24}}$

$s_p=36.144$

Confidence interval $=\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\alpha /2}·s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$

Confidence interval$=(9.0-1.6)\pm 1.711\times 36.144\sqrt{\frac{1}{14}+\frac{1}{12}}$

Confidence interval $=7.4\pm 24.33$

Confidence interval $=-16.92to31.72$