Question

According to a government agency, 54.1% of mothers in a certain region were breast-feeding their six-month-old babies in 2016. their children at six months in 2018. A random sample of 239 mothers with six-month-old babies was selected, 134 of which were a. Construct a 90% confidence interval to estimate the actual proportion of mothers with six-month-old babies that were breast-fee The confidence interval has a lower limit of and an upper limit of - (Round to three decimal places as needed.) b. What is the margin of error for this sample? The margin of error is a (Round to three decimal places as needed.) c. Is there evidence that this proportion has changed since 2016 based on this sample? This sample evidence that this proportion has changed since 2016, since the

Answer 1

Observed	Hypothesized
0.5607	0.541	p (as decimal)
134/239	129/239	p (as fraction)
134.	129.299	X
239	239	n
	0.0322	std. error
	0.61	z
	.5417	p-value (two-tailed)
	0.508	confidence interval 90.% lower
	0.613	confidence interval 90.% upper
	0.053	margin of error

a) lower Limit=0.508 and Upper limit= 0.613

b) Margin of error= 0.053

This sample has enough evidence that this proportion has changed since 2016, since the population proportion does not lies in above confidence interval.