Question

researchers conducted a study to determine whether magnets are effective in treating back pain. pain was measured using the visual analog scale, and the results shown below are among the results obtained in the study. higher scores correspond to greater pain levels. assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. complete parts (a) to (c) below.
reduction in pain level after magnet treatment ($\mu_1$): $n = 20$, $\bar{x} = 0.42$, $s = 1.01$
reduction in pain level after sham treatment ($\mu_2$): $n = 20$, $\bar{x} = 0.35$, $s = 1.56$

1. use a 0.01 significance level to test the claim that those treated with magnets have a greater mean reduction in pain than those given a sham treatment (similar to a placebo).

what are the null and alternative hypotheses?
a. $h_0: \mu_1 \
eq \mu_2$; $h_1: \mu_1 < \mu_2$
b. $h_0: \mu_1 = \mu_2$; $h_1: \mu_1 > \mu_2$
c. $h_0: \mu_1 < \mu_2$; $h_1: \mu_1 \geq \mu_2$
d. $h_0: \mu_1 = \mu_2$; $h_1: \mu_1 \
eq \mu_2$
the test statistic, $t$, is 0.17. (round to two decimal places as needed.)
the p - value is 0.433. (round to three decimal places as needed.)
state the conclusion for the test.
fail to reject the null hypothesis. there is not sufficient evidence to support the claim that those treated with magnets have a greater mean reduction in pain than those given a sham treatment.
b. construct a confidence interval appropriate for the hypothesis test in part (a)
$< \mu_1 - \mu_2 < $
(round to two decimal places as needed.)

Explanation:

Step1: Identify Hypotheses

The claim is that magnet - treated have greater mean pain reduction than sham - treated. Let $\mu_1$ be the mean for magnet treatment, $\mu_2$ for sham. So null hypothesis $H_0:\mu_1 = \mu_2$, alternative $H_1:\mu_1>\mu_2$, which matches option B.

Step2: Confidence Interval Formula

For two - sample t - interval (equal variances assumed), the formula is $(\bar{x}_1-\bar{x}_2)\pm t_{\alpha/2,df}\sqrt{s_p^2(\frac{1}{n_1}+\frac{1}{n_2})}$, where $s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}$, $n_1 = n_2=20$, $\bar{x}_1 = 0.42$, $\bar{x}_2 = 0.35$, $s_1 = 1.01$, $s_2 = 1.56$.

First, calculate $s_p^2$:
$(n_1 - 1)s_1^2=(20 - 1)\times1.01^2=19\times1.0201 = 19.3819$
$(n_2 - 1)s_2^2=(20 - 1)\times1.56^2=19\times2.4336 = 46.2384$
$n_1 + n_2-2=20 + 20-2 = 38$
$s_p^2=\frac{19.3819 + 46.2384}{38}=\frac{65.6203}{38}\approx1.7268$

Then, $\sqrt{s_p^2(\frac{1}{n_1}+\frac{1}{n_2})}=\sqrt{1.7268\times(\frac{1}{20}+\frac{1}{20})}=\sqrt{1.7268\times\frac{1}{10}}=\sqrt{0.17268}\approx0.4155$

$\bar{x}_1-\bar{x}_2=0.42 - 0.35 = 0.07$

For $\alpha = 0.01$ (two - tailed for confidence interval, since we are doing a one - tailed test in hypothesis, but confidence interval for the difference $\mu_1-\mu_2$), $df = 38$, $t_{\alpha/2,df}=t_{0.005,38}\approx2.712$

Margin of error $E = 2.712\times0.4155\approx1.127$

Confidence interval: $0.07\pm1.127$, so $(- 1.06,1.19)$ (rounded to two decimal places)

Answer:

a. The correct hypothesis option is B. $H_0:\mu_1=\mu_2$, $H_1:\mu_1 > \mu_2$
b. The confidence interval for $\mu_1-\mu_2$ is $\boldsymbol{-1.06<\mu_1 - \mu_2<1.19}$