Question

Example 14.1 (page 360) described NHANES survey data on the body mass index (BMI) of 654 young women. The mean BMI in the sample was x¯¯¯ = 26.8. We treated these data as an SRS from a Normally distributed population with standard deviation σ = 7.5. (a) Give three confidence intervals for the mean BMI μ in this population, using 90%, 95%, and 99% confidence. (b) What are the margins of error for 90%, 95%, and 99% confidence? How does increasing the confidence level change the margin of error of a confidence interval when the sample size and population standard deviation remain the same?

Answer 1

Answer:

Step-by-step explanation:

Hello!

The study variable is

X: body mass index (BMI) of a woman.

The population is normally distributed and has a standard deviation of σ= 7.5

There was a sample of n= 654 young women taken and a sample mean of X[bar]= 26.8 was estimated.

a.

To construct the asked CI, considering that the variable of interest is normally distributed and the population variance is known, the propper statistic to use is a standard normal $Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~N(0;1)$

The general structure of the confidence interval for the population mean is "point estimation" ± "margin of error"

Using the standard normal the formula to calculate the CI is:

X[bar]±$Z_{1-\alpha /2} * \frac{Sigma}{\sqrt{n} }$

Note that the only value that will change for all three intervals is the Z value:

90% Confidence interval

1-α= 0.90 ⇒ α=0.1

$Z_{1-\alpha /2} = Z_{1-(0.1/2)}= Z_{0.95}= 1.648$

26.8±$1.648 * \frac{7.5}{\sqrt{654} }$

[26.3167;27.2833]

95% Confidence interval

1-α= 0.95 ⇒ α=0.05

$Z_{1-\alpha /2}= Z_{1-(0.05/2)}= Z_{0.975}= 1.965$

26.8±$1.965* \frac{7.5}{\sqrt{654} }$

[26.2237;27.3763]

99% Confidence interval

1-α= 0.99 ⇒ α=0.01

$Z_{1-\alpha /2}= Z_{1-(0.01/2)}= Z_{0.995}= 2.586$

26.8±$2.586*\frac{7.5}{\sqrt{654} }$

[26.0416;27.5584]

b.

The margin of error is the semiamplitude (d) of the confidence interval, you calculate it you can calculate the margin of error by:

$d= \frac{Upperbond - Lowerbond}{2}$

90% Confidence interval

[26.3167;27.2833]

$d= \frac{27.2833-26.3167}{2}= 0.4833$

95% Confidence interval

[26.2237;27.3763]

$d= \frac{27.3763-26.2237}{2}= 0.5763$

99% Confidence interval

[26.0416;27.5584]

$d= \frac{27.5584-26.0416}{2}= 0.7584$

As you can see when comparing the margin of error of the three intervals, when you increase the confidence level of an interval, leaving all other data unchanged, the amplitude and margin of error of the interval increases. This is because there is a direct relationship between the confidence level and the amplitude of the confidence interval.

I hope it helps!