Use the following data to compute a 98% upper confidence bound for μ1 − μ2: m = 41 x = 42,700 s1 = 2030 n = 41 y = 36,275 s2 = 1360.

Question

asked Jan 25, 2021 100k views

1 Answer

Jason Kuhrt · Answer 1 · 2021-01-30T21:05:52+0000

Answer:

$(42700- 36275) -2.374 \sqrt{(2030^2)/(41) +(1360^2)/(41)}= 5519.071$

$(42700- 36275) +2.374 \sqrt{(2030^2)/(41) +(1360^2)/(41)}=7330.929$

Explanation:

For this case we have the following info given:

$n_1 = 41 , \bar X_1 =42700 , s_1 = 2030$

$n_2 = 41 , \bar X_2 =36375 , s_2 = 1360$

And for this case we want a 98% confidence interval. The significance would be:

$\alpha= 1-0.98=0.02$

The degrees of freedom are:

df = n_1 +n_2 -2= 41+41 -2= 80

And the critical value for this case is:

$t_(\alpha/2)= 2.374$

And the confidence interval would be given by:

$(\bar X_1 -\bar X_2) \pm t_(\alpha/2) \sqrt{(s^2_1)/(n_1)+(s^2_2)/(n_2)}$

And replacing we got:

$(42700- 36275) -2.374 \sqrt{(2030^2)/(41) +(1360^2)/(41)}= 5519.071$

$(42700- 36275) +2.374 \sqrt{(2030^2)/(41) +(1360^2)/(41)}=7330.929$