Question

a study was done using a treatment group and a placebo group. the results are shown in the table. 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) and (b) below. use a 0.05 significance level for both parts.

a. test the claim that the two samples are from populations with the same mean.
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
eq mu_2 )
c. ( h_0: mu_1 = mu_2 )
( h_1: mu_1 > mu_2 )
d. ( h_0: mu_1 < mu_2 )
( h_1: mu_1 geq mu_2 )
the test statistic, ( t ), is \\( \square \\). (round to two decimal places as needed.)
the p - value is \\( \square \\). (round to three decimal places as needed.)
state the conclusion for the test.
a. reject the null hypothesis. there is sufficient evidence to warrant rejection of the claim that the two samples are from populations with the same mean.
b. fail to reject the null hypothesis. there is not sufficient evidence to warrant rejection of the claim that the two samples are from populations with the same mean.
c. fail to reject the null hypothesis. there is sufficient evidence to warrant rejection of the claim that the two samples are from populations with the same mean.
d. reject the null hypothesis. there is not sufficient evidence to warrant rejection of the claim that the two samples are from populations with the same mean.
b. construct a confidence interval suitable for testing the claim that the two samples are from populations with the same mean.
\\( \square < mu_1 - mu_2 < \square \\)
(round to three decimal places as needed.)

treatment | \\( \mu \\) | n | \\( \bar{x} \\) | s
--- | --- | --- | --- | ---
\\( \mu_1 \\) | | 27 | 2.35 | 0.98
\\( \mu_2 \\) | | 30 | 2.62 | 0.69

Explanation:

Step 1: Identify Hypotheses

The claim is that the two populations have the same mean, so the null hypothesis (\(H_0\)) is \( \mu_1 = \mu_2 \). The alternative hypothesis (\(H_1\)) for a two - tailed test (since we are testing if they are the same, the opposite is they are different) is \( \mu_1
eq\mu_2 \). So the correct option is B: \(H_0:\mu_1 = \mu_2\), \(H_1:\mu_1
eq\mu_2\)

Step 2: Calculate Test Statistic (Pooled - t test)

First, we need to calculate the pooled variance \(s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}\)
Given \(n_1 = 27\), \(s_1 = 0.98\), \(n_2=30\), \(s_2 = 0.69\)
\((n_1 - 1)s_1^2=(27 - 1)\times(0.98)^2=26\times0.9604 = 24.9704\)
\((n_2 - 1)s_2^2=(30 - 1)\times(0.69)^2=29\times0.4761=13.8069\)
\(n_1 + n_2-2=27 + 30-2 = 55\)
\(s_p^2=\frac{24.9704 + 13.8069}{55}=\frac{38.7773}{55}\approx0.7050\)
\(s_p=\sqrt{0.7050}\approx0.8396\)

The test statistic \(t=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\)
Since \(H_0:\mu_1=\mu_2\), \(\mu_1-\mu_2 = 0\), \(\bar{x}_1 = 2.35\), \(\bar{x}_2=2.62\)
\(t=\frac{(2.35 - 2.62)-0}{0.8396\sqrt{\frac{1}{27}+\frac{1}{30}}}\)
\(\frac{1}{27}+\frac{1}{30}=\frac{30 + 27}{27\times30}=\frac{57}{810}\approx0.0704\)
\(\sqrt{0.0704}\approx0.2653\)
\(s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}=0.8396\times0.2653\approx0.2228\)
\(t=\frac{- 0.27}{0.2228}\approx - 1.21\) (rounded to two decimal places)

Step 3: Calculate P - value

The degrees of freedom \(df=n_1 + n_2-2=55\). Since this is a two - tailed test, the P - value is \(2\times P(t < - 1.21)\) (using t - distribution table or calculator)
Using a t - distribution calculator, \(P(t < - 1.21)\approx0.115\), so the P - value \(=2\times0.115 = 0.230\) (rounded to three decimal places)

Step 4: State Conclusion

The significance level \(\alpha = 0.05\). Since the P - value (\(0.230\))\(>\alpha(0.05)\), we fail to reject the null hypothesis. So the conclusion is: Fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that the two samples are from populations with the same mean. So the correct option is C.

Step 5: Confidence Interval

The formula for the confidence interval for \(\mu_1-\mu_2\) is \((\bar{x}_1-\bar{x}_2)\pm t_{\alpha/2,df}s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\)
\(\bar{x}_1-\bar{x}_2=2.35 - 2.62=-0.27\)
\(t_{\alpha/2,df}\) for \(\alpha = 0.05\) and \(df = 55\) is approximately \(2.004\)
\(s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\approx0.2228\) (from step 2)
Margin of error \(E = 2.004\times0.2228\approx0.446\)
The confidence interval is \(-0.27\pm0.446\), which is \((-0.716,0.176)\) (rounded to three decimal places)

Answer:

a. The null and alternative hypotheses: B. \(H_0:\mu_1 = \mu_2\), \(H_1:\mu_1
eq\mu_2\)
Test statistic: \(-1.21\)
P - value: \(0.230\)
Conclusion: C. Fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that the two samples are from populations with the same mean.
b. Confidence interval: \(-0.716<\mu_1-\mu_2<0.176\)