91影视

To determine the sample mean and standard deviation.

Since,$n=40$.

And the $95\%$confidence interval is$430$to $470$minutes.

Determine the sample mean falls in the middle of the confidence interval as:
$\stackrel{炉}{x}=\frac{430+470}{2}$
$=450$

Determine the margin of error as follows:

Margin of error$=450-430$

$=20$

Then, using table B and the degree of freedom as follows, get the critical value as:

$df=40-1$

$=39>30$

$t_{伪/2}=t_{0.025}$

$=2.042$

A confidence interval's margin of error for the population mean as:

Margin of error $=t_{伪/2}脳\frac{s}{\sqrt{n}}$

$=2.042脳\frac{s}{\sqrt{40}}$

$=0.3229s$

Finally, from the above:

$20=0.3229s$

$鈬�s=\frac{20}{0.3229}$

$=61.9447$

As a result, the mean is $450$ and standard deviation is $61.9447$.

A $95\%$chance that it will last between $430$and $470$ minutes. To comment on this interpretation.

Since the battery will either last between $430$ and $470$ minutes or it will not, the likelihood is either $0$or $1$.
A more accurate translation would be: We are $95\%$ confident that the average battery life time is between $430$ and $470$ minutes.
As a result, this is incorrect application.

To explain that agree with there is a $95\%$probability that the mean life time will fall between $430$and $470$ minutes.

No, since if the sample mean (on which the confidence interval is based) is extremely improbable to occur by chance for a given population mean, the likelihood of an alternative sample mean being in the same confidence interval is less than $95$percent.

As a result, no, a confidence interval provides a statement about an unknown population mean, not another sample mean.

To find that statistically correct interpretation of the confidence interval that could be published in a newspaper report.

Let, $n=40$.

Then, $95\%$confidence interval is $430$to $470$minutes
The standard deviation is $61.9447$, while the mean is $450$.
Hence, a statistically correct interpretation of the confidence interval that might be stated in a newspaper story might be that the population mean is between $430$and $470$ minutes.