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test the claim about the difference between two population means \\( \\…

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 \\) a. \\( h _ { 0 } : \mu _ { 1 } < \mu _ { 2 } \\)
\\( h _ { a } : \mu _ { 1 } \geq \mu _ { 2 } \\)
\\( \bigcirc \\) b. \\( h _ { 0 } : \mu _ { 1 } = \mu _ { 2 } \\)
\\( h _ { a } : \mu _ { 1 } \
eq \mu _ { 2 } \\)
\\( \bigcirc \\) c. \\( h _ { 0 } : \mu _ { 1 } \leq \mu _ { 2 } \\)
\\( h _ { a } : \mu _ { 1 } \gt \mu _ { 2 } \\)
\\( \bigcirc \\) d. \\( h _ { 0 } : \mu _ { 1 } > \mu _ { 2 } \\)
\\( h _ { a } : \mu _ { 1 } \leq \mu _ { 2 } \\)
\\( \bigcirc \\) e. \\( h _ { 0 } : \mu _ { 1 } \
eq \mu _ { 2 } \\)
\\( h _ { a } : \mu _ { 1 } = \mu _ { 2 } \\)
\\( \bigcirc \\) f. \\( h _ { 0 } : \mu _ { 1 } \geq \mu _ { 2 } \\)
\\( h _ { a } : \mu _ { 1 } < \mu _ { 2 } \\)
find the standardized test statistic t.
\\( t = \\)

Explanation:

Step1: Recall the formula for the t - statistic for two - sample t - test (unequal variances)

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

Step2: Substitute the given values into the formula

We are given \(\bar{x}_{1} = 2409\), \(\bar{x}_{2}=2300\), \(s_{1}=174\), \(s_{2} = 53\), \(n_{1}=14\), and \(n_{2}=10\)

First, calculate the numerator: \(\bar{x}_{1}-\bar{x}_{2}=2409 - 2300=109\)

Then, calculate the denominator:
\(\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}=\frac{174^{2}}{14}+\frac{53^{2}}{10}\)
\(=\frac{30276}{14}+\frac{2809}{10}\)
\( = 2162.5714+280.9\)
\(=2443.4714\)
\(\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}=\sqrt{2443.4714}\approx49.43\)

Now, \(t=\frac{109}{49.43}\approx2.205\)

Answer:

\(t\approx2.21\)