Question

There was concern among health officials in the Windham area that an unusually large percentage of newborn babies had abnormally low birth weights (less than 5.5 lbs.). A sample of 250 newborns showed 20 of these babies had abnormally low birth weight.

Construct a 95% confidence interval for the true proportion of babies in the Windham area who have abnormally low birth weight.

	a.	.10 ± .042
	b.	.10 ± .042
	c.	.08 ± .028
	d.	.08 ± .034
	e.	.08 ± .072

Suppose that nationally the percentage of babies born with low birth weight is 4%. Would the health officials in the Windham be justified in their concern – that the number of babies born with low birth weight in Windham is unusually large?

	a.	Yes, since the national rate of 4% falls below the 95% confidence interval for the population proportion for the Windham area.
	b.	No, even though the national rate of 4% falls outside the confidence interval constructed for the Windham data, we should conclude that the Windham rate is not significantly higher.
	c.	No, the national rate of 4% falls within the confidence interval for p, the population proportion for the Windham area.
	d.	No conclusion can be drawn since the correct confidence interval would be a confidence interval for µ, since the question involves birth weight.
	e.	Yes, even though the national rate of 4% falls within the confidence interval for the Windham data, it is still correct to conclude that the Windham rate is significantly higher than the national rate.

Suppose the health official would like to do a follow-up study next year. She would like to continue using a 95% confidence interval but she would like to reduce the margin of error to 2.5%. How large a sample would she need for this follow-up study? To calculate the necessary sample size, assume that she will use her estimate of p, the population proportion, from her current study.

	a.	453
	b.	525
	c.	301
	d.	618
	e.	750

Answer 1

zα/2 = 1.9599

Lower Bound = p̂ - zα/2•√p̂(1 - p̂)/n

= 0.08 - (1.9599639)(0.01715805)

= 0.08-0.034

= 0.0463707645

Upper Bound = p̂ + zα/2•√p̂(1 - p̂)/n

= 0.08 + (1.9599)(0.017158)

= 0.08+0.034

= 0.1136292355

Confidence Interval = (0.0463, 0.113)

#confidence interval is from 4.6% to 11.3%

option a is correct. yes because 4% falls below the confidence interval.

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