Question

In the U.S., newborn babies weight an average of 7.2 pounds (μ). The standard deviation is 1.22 pounds (σ). The population distribution of birth weights is roughly Normal. You draw a simple random sample of 60 births from the population and find that the sample mean is 7.4 pounds. NOTE: Use the most precise estimate of z or z* you can in these problems as in all your written results. Draw a picture of the distribution, label, and shade as appropriate (this is not required but will be very helpful.) (a) Between what two values (in pounds) will 95% of the population fall? (b) Between what two values (in pounds) will 95% of sample means fall? (c) Are confidence intervals calculated using x-bar or mu? (d) Keeping your answer to part (c) in mind, calculate a 95% confidence interval for the mean birth weight of babies in the U.S. Give a “long-hand” interpretation of the confidence interval.

Answer 1

a) P(-x < x < x) = 0.95

or, P(-z < Z < z) = 0.95

or, P(Z < z) - P(Z < -z) = 0.95

Or, P(Z < z) - (1 - P(Z < z)) = 0.95

or, 2P(Z < z) = 1.95

or, P(Z < z) = 0.975

or, z = 1.96

or, (x - 7.2)/1.22 = 1.96

or, x = 1.96 * 1.22 + 7.2

or, x = 9.59

So the other value is = 7.2 - (9.59 - 7.2) = 4.81

b) P(-x < x < x) = 0.95

or, P(-z < Z < z) = 0.95

or, P(Z < z) - P(Z < -z) = 0.95

Or, P(Z < z) - (1 - P(Z < z)) = 0.95

or, 2P(Z < z) = 1.95

or, P(Z < z) = 0.975

or, z = 1.96

or, ( - 7.2)/(1.22/) = 1.96

or, = 1.96 * (1.22/) + 7.2

or, = 7.51

So the other value is = 7.2 - (7.51 - 7.2) = 6.89

c) Confidence intervals are calculated using .

d) At 95% confidence interval the critical value is z_0.025 = 1.96

The 95% confidence for population mean is

+/- z_0.025 *

= 7.4 +/- 1.96 * 1.22/

= 7.4 +/- 0.3087

= 7.0913, 7.7087

We are 95% confidence that the true population mean of babies weight lies in the above interval .