Respuesta :
The equation of this would be
[tex]true \ mean=mean \ +/- \ z\frac{s}{ \sqrt{n} } [/tex]
The z-value for a 95% confidence level is equal to 1.96.
Then the lower limit would be:
[tex]85-1.96 \frac{12}{ \sqrt{6} }=75.398 [/tex]
And the higher limit would be:
[tex]85+1.96 \frac{12}{ \sqrt{6} }=94.60[/tex]
Therefore, the answer is
c)75.4 to 94.6
I hope I was able to explain it clearly. Have a good day :)
[tex]true \ mean=mean \ +/- \ z\frac{s}{ \sqrt{n} } [/tex]
The z-value for a 95% confidence level is equal to 1.96.
Then the lower limit would be:
[tex]85-1.96 \frac{12}{ \sqrt{6} }=75.398 [/tex]
And the higher limit would be:
[tex]85+1.96 \frac{12}{ \sqrt{6} }=94.60[/tex]
Therefore, the answer is
c)75.4 to 94.6
I hope I was able to explain it clearly. Have a good day :)
The correct answer is:
c)75.4 to 94.6
Explanation:
The formula for a confidence interval is:
[tex] \mu \pm z*(\frac{\sigma}{\sqrt{n}}) [/tex],
where μ is the mean, z is the z-score associated with the level of confidence we want, σ is the standard deviation, and n is the sample size.
Our mean is 85, our standard deviation is 12, our sample size is 6, and since we want 95% confidence, our z-score is 1.96:
[tex] 85\pm 1.96(\frac{12}{\sqrt{6}})=85\pm 9.6=85-9.6, 85+9.6=75.4, 94.6 [/tex]