Mathematics, 26.03.2020 22:20 asilverman

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.)

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Answer from: tclangley

90% confidence

$28.9-1.64\frac{7.2}{\sqrt{651}}=28.44$

$28.9+1.64\frac{7.2}{\sqrt{651}}=29.36$

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

95% confidence

$28.9-1.96\frac{7.2}{\sqrt{651}}=28.35$

$28.9+1.96\frac{7.2}{\sqrt{651}}=29.45$

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

99% confidence

$28.9-2.58\frac{7.2}{\sqrt{651}}=28.17$

$28.9+2.58\frac{7.2}{\sqrt{651}}=29.63$

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

Step-by-step explanation:

Previous concepts

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".

$\bar X=28.9$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=7.2$ 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:

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

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , 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 $t_{\alpha/2}=1.64$

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

$28.9-1.64\frac{7.2}{\sqrt{651}}=28.44$

$28.9+1.64\frac{7.2}{\sqrt{651}}=29.36$

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 $\alpha=0.05$ and $\alpha/2 =0.025$ , 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 $t_{\alpha/2}=1.96$

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

$28.9-1.96\frac{7.2}{\sqrt{651}}=28.35$

$28.9+1.96\frac{7.2}{\sqrt{651}}=29.45$

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 $\alpha=0.01$ and $\alpha/2 =0.005$ , 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 $z_{\alpha/2}=2.58$

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

$28.9-2.58\frac{7.2}{\sqrt{651}}=28.17$

$28.9+2.58\frac{7.2}{\sqrt{651}}=29.63$

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

Answer from: Quest

We need more info then that

Answer from: Quest

160

step-by-step explanation:

idk

i hope i

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We have the survey data on the body mass index (BMI) of 651 young women. The mean BMI in the sample...

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