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 } = \mu _ { 2 } ; \alpha = 0.10 \\). assume \\( \sigma _ { 1 } ^ { 2 } = \sigma _ { 2 } ^ { 2 } \\)
sample statistics: \\( \overline { x } _ { 1 } = 31.4, s _ { 1 } = 3.4, n _ { 1 } = 11 \\) and
\\( \overline { x } _ { 2 } = 33.9, s _ { 2 } = 2.4, n _ { 2 } = 19 \\)
identify the null and alternative hypotheses. choose the correct answer below.
\\( \bigcirc \\) a. \\( h _ { 0 } : \mu _ { 1 } \leq \mu _ { 2 } \\)
\\( h _ { a } : \mu _ { 1 } > \mu _ { 2 } \\)
\\( \bigcirc \\) b. \\( h _ { 0 } : \mu _ { 1 } < \mu _ { 2 } \\)
\\( h _ { a } : \mu _ { 1 } \geq \mu _ { 2 } \\)
\\( \bigcirc \\) c. \\( h _ { 0 } : \mu _ { 1 } > \mu _ { 2 } \\)
\\( h _ { a } : \mu _ { 1 } \leq \mu _ { 2 } \\)
\\( \bigcirc \\) d. \\( h _ { 0 } : \mu _ { 1 } = \mu _ { 2 } \\)
\\( h _ { a } : \mu _ { 1 } \
eq \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 = \\) (round to two decimal places as needed.)

Explanation:

Step1: Calculate the pooled variance \(s_p^2\)

The formula for pooled variance when \(\sigma_1^2=\sigma_2^2\) is \(s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}\)
Substitute \(n_1 = 11\), \(s_1 = 3.4\), \(n_2 = 19\), \(s_2 = 2.4\)
\(s_p^2=\frac{(11 - 1)\times3.4^2+(19 - 1)\times2.4^2}{11 + 19-2}=\frac{10\times11.56+18\times5.76}{28}=\frac{115.6 + 103.68}{28}=\frac{219.28}{28}\approx7.83\)

Step2: Calculate the standardized test statistic \(t\)

The formula for \(t\) - statistic is \(t=\frac{\bar{x}_1-\bar{x}_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\)
We know \(\bar{x}_1 = 31.4\), \(\bar{x}_2 = 33.9\), \(s_p=\sqrt{7.83}\approx2.8\), \(n_1 = 11\), \(n_2 = 19\)
\(\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}=\sqrt{\frac{19 + 11}{11\times19}}=\sqrt{\frac{30}{209}}\approx0.38\)
\(t=\frac{31.4 - 33.9}{2.8\times0.38}=\frac{- 2.5}{1.064}\approx - 2.35\)

Answer:

\(t\approx - 2.35\)