Question

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Answer 1

$\bar x1$ = 13.25

s1 = 4.5036

 $\bar x2$ = 16.25

s2 = 2.1373

Option - D) $H_{0}: \mu1 = \mu2$

  $H_{1} : \mu1 < \mu2$

The test statistic is

$t = \frac{\bar x1 - \bar x2}{\sqrt{\frac{s1^2}{n1} + \frac{s2^2}{n2}}}$

  $= \frac{13.25 - 16.25}{\sqrt{\frac{(4.5036)^2}{24} + \frac{(2.1373)^2}{12}}}$

$df = \frac{(s1^2/n1 + s2^2/n2)^2}{\frac{(s1^2/n1)^2}{n1 - 1} + \frac{(s2^2/n2)^2}{n2 - 1}}$

  $= \frac{((4.5036)^2/24 + (2.1373)^2/12)^2}{\frac{((4.5036)^2/24)^2}{23} + \frac{((2.1373)^2/12)^2}{11}}$

  $=33$

P-value = P(T < -2.71)

= 0.0034

Since the P-value is less than the significance level (0.0034 < 0.1), so we should reject the null hypothesis.

Option - C) Reject the null hypothesis. There is sufficient evidence to support the claim that Archbishap have lower mean longetivity.