We have the survey data on the body mass index (BMI) of 651 young women. The mean BMI in the sample was x = 28.9. We treated these data as an SRS from a Normally distributed population with standard deviation σ = 7.2. Give confidence intervals for the mean BMI and the margins of error for 90%, 95%, and 99% confidence. (Round your answers to two decimal places.)

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".

[tex]\bar X=28.9[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

[tex]\sigma=7.2[/tex] represent the population standard deviation

n=651 represent the sample size

90% confidence

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

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]t_{\alpha/2}=1.64[/tex]

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

[tex]28.9-1.64\frac{7.2}{\sqrt{651}}=28.44[/tex]

[tex]28.9+1.64\frac{7.2}{\sqrt{651}}=29.36[/tex]

So on this case the 90% confidence interval would be given by (28.44;29.36)

95% confidence

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that [tex]t_{\alpha/2}=1.96[/tex]

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

[tex]28.9-1.96\frac{7.2}{\sqrt{651}}=28.35[/tex]

[tex]28.9+1.96\frac{7.2}{\sqrt{651}}=29.45[/tex]

So on this case the 95% confidence interval would be given by (28.35;29.45)

99% confidence

Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that [tex]z_{\alpha/2}=2.58[/tex]

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

[tex]28.9-2.58\frac{7.2}{\sqrt{651}}=28.17[/tex]

[tex]28.9+2.58\frac{7.2}{\sqrt{651}}=29.63[/tex]

So on this case the 99% confidence interval would be given by (28.17;29.63)