We have to calculate the 90% confidence interval for the mean.
The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value
(NOTE: although we have a relatively big sample, so it wouldn't be too wrong to approximate the standard deviation of the population with the sample one.. Even though, we will use the Student's t test as usual for unknown population standard deviations).
When σ is not known, s divided by the square root of N is used as an estimate of σM:
[tex]s_M=\dfrac{s}{\sqrt{N}}=\dfrac{7.5}{\sqrt{245}}=\dfrac{7.5}{15.652}=0.479[/tex]
The degrees of freedom for this sample size are:
[tex]df=n-1=245-1=244[/tex]
The t-value for a 90% confidence interval and 244 degrees of freedom is t = 1.651.
The margin of error (MOE) can be calculated as:
[tex]MOE=t\cdot s_M=1.651\cdot0.479=0.791[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]\begin{gathered} LL=M-t\cdot s_M=28.6-0.791=27.8 \\ UL=M+t\cdot s_M=28.6+0.791=29.4 \end{gathered}[/tex]
The confidence interval for the population mean is:
[tex]27.8<\mu<29.4[/tex]
If we compare this interval with the one calcualted for a smaller sample, we can see that the limits are very similar and both CI contain both sample means.
Answer:
27.8 hg < μ < 29.4 hg.
A. No, because the confidence interval limits are similar.
