Question

Here are summary statistics for randomly selected weights of newborn​ girls: nequals=174174​, x overbarxequals=30.930.9 ​hg, sequals=7.57.5 hg. Construct a confidence interval estimate of the mean. Use a 9595​% confidence level. Are these results very different from the confidence interval 29.629.6 hgless than

Answer 1

The 95% confidence interval would be given by (29.780;32.020)

Explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X=30.9` represent the sample mean for the sample

`\mu` population mean (variable of interest)

s=7.5 represent the sample standard deviation

n=174 represent the sample size

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=174-1=173`

Since the Confidence is 0.95 or 95%, the value of
`\alpha=0.05` and
`\alpha/2 =0.025`, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,173)".And we see that
`t_(\alpha/2)=1.97`, this value is similar to the obtained with the normal standard distribution since the sample size is large to approximate the t distribution with the normal distribution.

Now we have everything in order to replace into formula (1):

`30.9-1.97(7.5)/(√(174))=29.780`

`30.9+1.97(7.5)/(√(174))=32.020`

So on this case the 95% confidence interval would be given by (29.780;32.020)

The value 29.6 is not contained on the interval calculated.