Question

We have the survey data on the body mass index (BMI) of 642 young women. The mean BMI in the sample was x¯=26.1. We treated these data as an SRS from a Normally distributed population with standard deviation ?=7.3.

Give confidence intervals for the mean BMI and the margins of error for 90%, 95%, and 99% confidence.

Conf. Level	Interval (±±0.01)	margins of error (±±0.0001)
90%	to
95%	to
99%	to

Answer 1

Solution :

Given that,

$\bar x$ = 26.1

$\sigma$ = 7.3

n = 642

a ) At 90% confidence level the z is ,

$\alpha$  = 1 - 90% = 1 - 0.90 = 0.1

$\alpha$ / 2 = 0.1 / 2 = 0.05

Z_$\alpha$/2 = Z_0.05 = 1.645

Margin of error = E = Z_$\alpha$/2* ($\sigma$$\sqrt$/n)

= 1.645 * (7.3 / $\sqrt$642) = 0.47

At 90% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

26.1 - 0.47 < $\mu$ < 26.1 + 0.47

25.63< $\mu$ < 26.57

(25.63 , 26.57)

b ) At 95% confidence level the z is ,

$\alpha$  = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z_$\alpha$/2 = Z_0.025 = 1.960

Margin of error = E = Z_$\alpha$/2* ($\sigma$$\sqrt$/n)

= 1.960 * (7.3/ $\sqrt$642) = 0.56

At 95% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

26.1 - 0.56 < $\mu$ < 26.1 + 0.56

25.54 < $\mu$ < 26.66

(25.54, 26.66)

c )At 99% confidence level the z is ,

$\alpha$  = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$ / 2 = 0.01 / 2 = 0.005

Z_$\alpha$/2 = Z_0.005 = 2.576

Margin of error = E = Z_$\alpha$/2* ($\sigma$$\sqrt$/n)

= 2.576 * (7.3 / $\sqrt$642) = 0.74

At 99% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

26.1 - 0.74 < $\mu$ < 26.1 + 0.74

25.36 < $\mu$ < 26.84

(25.36, 26.84)