Question

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1. A simple random sample of birth weights in the United States has a mean of 3433g. The standard deviation of all birth weights (for the population) is 495g.
   1. If this data came from a sample size of 75, construct a 95% confidence interval estimate of the mean birth weight in the United States.

2. If this data came from a sample size of 7,500, construct a 95% confidence interval estimate of the mean birth weight in the United States.

3. Which of the confidence levels above is wider? Why?

Answer 1

Solution :

Given that,

= 3433

= 495

n = 75

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_/2* ( /n)

=1.96 * (495 / 75)

= 112.0290

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

- E < < + E

3433 - 112.0290 < < 3433 + 112.0290

3320.9710< < 3545.0290

(3320.9710< < 3545.0290 )

Solution :

Given that,

= 3433

= 495

n = 75

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_/2* ( /n)

=1.96 * (495 / 7500)

=1120.2905

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

- E < < + E

3433 - 1120.2905 < < 3433 + 1120.2905

2312.7095< < 4553.2905

(2312.7095 , 4553.2905 )

a ) is above because sample size is larger