Business statistics need help

Question

According to a government​ agency, 48.6​% of mothers in a certain region were​ breast-feeding their four​-month-old

babies in 2016. Suppose a health insurance company would like to estimate the proportion of mothers in the region who​ breast-fed their children at four months in 2018. A random sample of 226 mothers with four​-month-old babies was​ selected, 127

of which were​ breast-feeding. Complete parts a through c.

____________________________________________________

a. Construct a 90​% confidence interval to estimate the actual proportion of mothers with four​-month-old

babies that were​ breast-feeding in

2018. 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 error for this sample?

C. Is there evidence that this proportion has changed since 2016 based on this sample? The sample (provides, or does not provide) evidence that this proportion has changed since 2016.

Answer 1

Step 1: State the Given Information

N=226, x =127, p= x/n =127/226 =562 q=1-.562=.438

p = .486 1 - alpha = .90 and alpha = .10

Step2: Verify Sampling Distribution of p^ can be normal

np^ >= 5 226(.562) >= 5 127 >=5 check

nq^ >=5 226(.438) >=5 99 >=5 check

Step 3: Find the critical values for z.

z_c = invnorm(.10/2)= -1.645 and z_c=invnorm(1-.10/2) =1.645.

Step 4: Find the margin of error E

E=z_c*sqr(p^*q^/n) = (1.64)sqr((.562*.438)/226)=.054

Step 5: Find the Lower and Upper Limit for CI.

p^ - E < p < p^ + E

.562 -.054 < p < p^ + +.054

.508 < p < .616

Step 6: State the CI.

90% CI

50.8% < p < 61.6%

The Margin of Error is 10%

There is evidence that his proportion has changed since 2016 based on this sample since the original p is less than the estimated p value range.