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Answered question

2021-02-04

Let two independent random samples, each of size 10, from two normal distributions

N (μ_{1}, σ_{2}) and N (μ_{2}, σ_{2})

yield

x = 4.8, s_{1}^{2}

= 8.64, y = 5.6, s_{2}^{2}

= 7.88.
Find a 95% confidence interval for

μ_{1} - μ_{2}

Answer & Explanation

l1koV

Skilled2021-02-05Added 100 answers

Step 1
$A 100 (1 - α) %$ confidence interval on the difference is
${\overset{―}{x}}_{1} - {\overset{―}{x}}_{2} - t_{\frac{α}{2}, n_{1} + n_{2} - 1} s_{p} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}} \leq μ_{1} - μ_{2} \leq {\overset{―}{x}}_{1} - {\overset{―}{x}}_{2}, + t_{\frac{α}{2}, n_{1} + n_{2} - 1} s_{p} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}$
where $s_{p} = \sqrt{\frac{(n_{1} - 1) s_{1}^{2} + (n_{2} - 1) s_{2}^{2}}{n_{1} + n_{2} - 1}}$
Step 2
Considering 95% confidence interval,
$α = 0.05$
${\overset{―}{x}}_{1} = 4.8$
${\overset{―}{x}}_{2} = 5.6$
$s_{1}^{2} = 8.64$
$s_{2}^{2} = 7.88$
$n_{1} = 10$
$n_{2} = 10$
Therefore
$t_{\frac{α}{2}, n_{1} + n_{2} - 1} = t_{\frac{0.05}{2}, 10 + 10 - 1}$
$= t_{0.025, 19}$
=2.093
Step 3
To calculate $s_{p}$
$s_{p} = \sqrt{\frac{(n_{1} - 1) s_{1}^{2} + (n_{2} - 1) s_{2}^{2}}{n_{1} + n_{2} - 1}}$
$s_{p} = (\frac{\sqrt{(10 - 1) 8.64 + (10 - 1) 7.88}}{19})$
$s_{p} = 2.797$
Step 4
Now, by substituting all the values in the equation, we get
${\overset{―}{x}}_{1} - {\overset{―}{x}}_{2} - t_{\frac{α}{2}, n_{1} + n_{2} - 1} s_{p} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}} \leq μ_{1} - μ_{2} \leq {\overset{―}{x}}_{1} - {\overset{―}{x}}_{2}$

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