Answer:
"0.0373" seems to be the appropriate solution.
Step-by-step explanation:
The given values are:
n₁ = 43
n₂ = 35
[tex]\bar{x_1}=217800[/tex]
[tex]\bar{x_2}=204700[/tex]
[tex]\sigma_1=30300[/tex]
[tex]\sigma_2=33800[/tex]
Now,
The test statistic will be:
⇒ [tex]Z=\frac{\bar{x_1}-\bar{x_2}}\sqrt{\frac{\sigma_1^2}{n_1} +\frac{\sigma_2^2}{n_2} }[/tex]
On substituting the given values in the above formula, we get
⇒ [tex]=\frac{217800-204400}{\sqrt{\frac{(30300)^2}{43} +\frac{(33800)^2}{35} } }[/tex]
⇒ [tex]=\frac{217800-204400}{\sqrt{\frac{918090000}{43} +\frac{1142440000}{35} } }[/tex]
⇒ [tex]=\frac{13400}{7347.93}[/tex]
⇒ [tex]=1.7828[/tex]
then,
P-value will be:
= [tex]P(Z>1.7828)[/tex]
= [tex]0.0373[/tex]
