Question

QUESTION PART A: Assume that a sample is used to estimate a population mean μμ. Find the 80% confidence interval for a sample of size 882 with a mean of 51.3 and a standard deviation of 17.1. Enter your answer as a tri-linear inequality accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

_____< μ <______

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

QUESTION PART B: Giving a test to a group of students, the grades and gender are summarized below

	A	B	C	Total
Male	8	14	16	38
Female	3	19	15	37
Total	11	33	31	75

Let pp represent the percentage of all male students who would receive a grade of A on this test. Use a 98% confidence interval to estimate p to three decimal places.

Enter your answer as a tri-linear inequality using decimals (not percents).

_____ < p <_____

Answer 1

A)

sample mean 'x̄=	51.30
sample size    n=	882.00
std deviation σ=	17.100
std errror ='σx=σ/√n=	0.5758

for 80 % CI value of z=		1.282
margin of error E=z*std error =		0.74
lower bound=sample mean-E=		50.562
Upper bound=sample mean+E=		52.038

80% confidence interval for the sample : 50.6 < μ < 52.0

B)

sample success x =		8
sample size          n=		38
sample proportion p̂ =x/n=		0.2105
std error se= √(p*(1-p)/n) =		0.0661
for 98 % CI value of z=		2.326
margin of error E=z*std error   =		0.1538
lower bound=p̂ -E                       =		0.0567
Upper bound=p̂ +E                     =		0.3643

98% confidence interval to estimate p = 0.0567 < p < 0.3643