Answer:
The correct answer is "771.44; 779.16".
Step-by-step explanation:
Given:
Number of samples,
[tex]n = 40[/tex]
Mean,
[tex]\bar x = 775.3[/tex]
Standard deviation,
[tex]\sigma = 14.9[/tex]
At 90% confidence interval,
[tex]\alpha = 1-0.90[/tex]
[tex]=0.10[/tex]
[tex]\frac{\alpha}{2} = 0.05[/tex]
From normal distribution at 90% confidence level,
[tex]Z_{\frac{\alpha}{2} } = 1.64[/tex]
Now,
Margin of error will be:
⇒ [tex]E=Z_{\frac{\alpha}{2} }\times \frac{\sigma}{\sqrt{n} }[/tex]
By substituting the values, we get
[tex]=1.64\times \frac{14.9}{\sqrt{40} }[/tex]
[tex]=3.86[/tex]
hence,
The mean at 90% CI will be:
= [tex]\bar X \pm \ E[/tex]
= [tex]775.3\ \pm \ 3.86[/tex]
= [tex]771.44; 779.16[/tex]
