Question

*without using interval functions on calculator*

26. Birth Weights A simple random sample of birth weights in the United States has a mean of 3433 g. The standard deviation of all birth weights is 495 g a. Using a sample size of 75, construct a 95% confidence interval estimate of the mean birth weight in the United States. b. Using a sample size of75,000, construct a 95% confidence interval estimate of the mean birth weight in the United States. c. Which of the preceding confidence intervals is wider? Why?

Answer 1

Solution(a)

Given in the question

Mean = 3433

Standard deviation = 495

If no.of sample = 75

Than 95% confidence interval can be calculated as

Mean +/- Zalpha/2*SD/Sqrt(n)

3433+/-1.96*495/Sqrt(75)

= 3433+/-112.03

So 95% confidence interval is

3320.97 to 3545.03

Solution(b)

If n=75000

3433+/-1.96*496/Sqrt(75000)

= 3433+/-3.54

So 95% confidence interval is

3429.46 to 3436.54

Solution:

As sample size is less than confidence interval is wider.

If sample size is 75 than confidence interval is wider because sample size is less.