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 0.10 significance level for both parts.

treatment placebo
μ μ₁ μ₂
n 35 30
\\( \overline { x } \\) 2.32 2.62
s 0.62 0.88

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 } > mu _ { 2 } \\)
○ c. \\( h _ { 0 } : mu _ { 1 } = mu _ { 2 } \\)
\\( h _ { 1 } : mu _ { 1 }
eq 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.)

Explanation:

Step1: Calculate the test statistic formula

The formula for the test statistic \(t\) for two - sample \(t\) - test (unequal variances) is \(t=\frac{\bar{x}_{1}-\bar{x}_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}\)

Given \(\bar{x}_{1} = 2.32\), \(s_{1}=0.62\), \(n_{1}=35\), \(\bar{x}_{2}=2.62\), \(s_{2}=0.88\), \(n_{2}=30\)

Step2: Substitute values into the formula

First, calculate the numerator: \(\bar{x}_{1}-\bar{x}_{2}=2.32 - 2.62=- 0.3\)

Then calculate the denominator:
\(\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}=\sqrt{\frac{0.62^{2}}{35}+\frac{0.88^{2}}{30}}\)
\(=\sqrt{\frac{0.3844}{35}+\frac{0.7744}{30}}\)
\(=\sqrt{0.010983 + 0.025813}\)
\(=\sqrt{0.036796}\approx0.1918\)

Step3: Calculate the \(t\) - value

\(t=\frac{-0.3}{0.1918}\approx - 1.56\)

Answer:

The test statistic \(t\) is \(-1.56\)