Question

5) Here are summary statistics for randomly selected weights of newborn girls: n=197, x=32.7hg, s=6.7 hg. Construct a confidence interval estimate of the mean. Use a 98% confidence level. Are these results very different from the confidence interval 31.6 hg< μ < 34.8 hg with only 12 sample values, x=33.2 hg, and s=2.5 hg? What is the confidence interval for the population mean μ?

3) The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first determine the percentage of adults who have heard of the brand. How many adults must he survey in order to be 80% confident that his estimate is within five percentage points of the true population​ percentage?

Complete parts​ (a) through​ (c) below.

A) Assume that nothing is known about the percentage of adults who have heard of the brand.
a.n=_________​(Round up to the nearest​ integer.)

B) Assume that a recent survey suggests that about 85​% of adults have heard of the brand.

b.n=_________(Round up to the nearest​ integer.)

Answer 1

Solution :

5) Given that,

Point estimate = sample mean = = 32.7 hg

sample standard deviation = s = 6.7 hg

sample size = n = 197

Degrees of freedom = df = n - 1 = 197 - 1 = 196

At 98% confidence level

= 1 - 98%

=1 - 0.98 =0.02

/2 = 0.01

t/2,df = t0.01,196 = 2.346

Margin of error = E = t/2,df * (s /n)

= 2.346 * ( 6.7 / 197)

Margin of error = E = 1.1

The 98% confidence interval estimate of the population mean is,

- E < < + E

32.7 - 1.1 < < 32.7 + 1.1

31.6 hg < < 33.8 hg

Yes, because the confidence interval limits are not similar

3) Given that,

A) =  1 - = 0.5

margin of error = E = 0.05

At 80% confidence level

= 1 - 80%  

= 1 - 0.80 =0.20

/2 = 0.10

Z/2 = Z0.10  = 1.282

sample size = n = (Z / 2 / E )2 * * (1 - )

= (1.282 / 0.05)2 * 0.5 * 0.5

= 164.35

sample size = n = 165

B) = 0.85

1 - = 1 - 0.85 = 0.15

sample size = n = (Z / 2 / E )2 * * (1 - )

= (1.282 / 0.05)2 * 0.85 * 0.15

= 83.81

sample size = n = 84