Question

test the claim about the difference between two population means \\( \mu _ { 1 } \\) and \\( \mu _ { 2 } \\) at the level of significance \\( \alpha \\). assume the samples are random and independent, and the populations are normally distributed.

claim: \\( \mu _ { 1 } \leq \mu _ { 2 } ; \alpha = 0.10 \\). assume \\( \sigma _ { 1 } ^ { 2 } \
eq \sigma _ { 2 } ^ { 2 } \\)

sample statistics: \\( \overline { x } _ { 1 } = 2409, s _ { 1 } = 174, n _ { 1 } = 14 \\) and
\\( \overline { x } _ { 2 } = 2300, s _ { 2 } = 53, n _ { 2 } = 10 \\)

identify the null and alternative hypotheses. choose the correct answer below.

\\( \bigcirc \mathrm { a } \\) \\( h _ { 0 } : \mu _ { 1 } < \mu _ { 2 } \\)
\\( h _ { a } : \mu _ { 1 } \geq \mu _ { 2 } \\)
\\( \bigcirc \mathrm { b } \\) \\( h _ { 0 } : \mu _ { 1 } = \mu _ { 2 } \\)
\\( h _ { a } : \mu _ { 1 } \
eq \mu _ { 2 } \\)
\\( \bigcirc \mathrm { c } \\) \\( h _ { 0 } : \mu _ { 1 } \leq \mu _ { 2 } \\)
\\( h _ { a } : \mu _ { 1 } > \mu _ { 2 } \\)
\\( \bigcirc \mathrm { d } \\) \\( h _ { 0 } : \mu _ { 1 } > \mu _ { 2 } \\)
\\( h _ { a } : \mu _ { 1 } \leq \mu _ { 2 } \\)
\\( \bigcirc \mathrm { e } \\) \\( h _ { 0 } : \mu _ { 1 } \
eq \mu _ { 2 } \\)
\\( h _ { a } : \mu _ { 1 } = \mu _ { 2 } \\)
\\( \bigcirc \mathrm { f } \\) \\( h _ { 0 } : \mu _ { 1 } \geq \mu _ { 2 } \\)
\\( h _ { a } : \mu _ { 1 } < \mu _ { 2 } \\)

find the standardized test statistic \\( t \\).

\\( t = 2.21 \\)
(round to two decimal places as needed.)

find the p - value.

\\( p = \square \\)
(round to three decimal places as needed)

Explanation:

Step1: Calculate the degrees of freedom

The formula for degrees of freedom when \(\sigma_{1}^{2}
eq\sigma_{2}^{2}\) is \(df=\frac{(\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}})^{2}}{\frac{(\frac{s_{1}^{2}}{n_{1}})^{2}}{n_{1}-1}+\frac{(\frac{s_{2}^{2}}{n_{2}})^{2}}{n_{2}-1}}\)
Substitute \(s_{1} = 174\), \(n_{1}=14\), \(s_{2}=53\), \(n_{2}=10\)
\(\frac{s_{1}^{2}}{n_{1}}=\frac{174^{2}}{14}=\frac{30276}{14} = 2162.5714\)
\(\frac{s_{2}^{2}}{n_{2}}=\frac{53^{2}}{10}=\frac{2809}{10}=280.9\)
\((\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}})^{2}=(2162.5714 + 280.9)^{2}=(2443.4714)^{2}=5969547.7\)
\(\frac{(\frac{s_{1}^{2}}{n_{1}})^{2}}{n_{1}-1}=\frac{2162.5714^{2}}{13}=\frac{4676739.1}{13}\approx359749.16\)
\(\frac{(\frac{s_{2}^{2}}{n_{2}})^{2}}{n_{2}-1}=\frac{280.9^{2}}{9}=\frac{78904.81}{9}\approx8767.20\)
\(df=\frac{5969547.7}{359749.16 + 8767.20}\approx16\)

Step2: Find the P - value

Since the test is a right - tailed test (because \(H_{a}:\mu_{1}>\mu_{2}\)) and \(t = 2.21\) with \(df\approx16\)
Using a t - distribution table or a calculator, the P - value is the probability of getting a t - statistic greater than \(2.21\) with \(df = 16\)
\(P\approx0.023\)

Answer:

\(P = 0.023\)