This document provides solutions to statistical estimation and confidence interval problems. It defines key statistical concepts like confidence level, margin of error, and t and chi-square distributions. Several multi-part problems are solved involving determining sample sizes needed, calculating point estimates and confidence intervals for means, proportions, variances and standard deviations using the relevant statistical formulas and distributions.

1. 1
Statistics Sample Test 3B (Part Two) Solution
Estimates & Sample Sizes
1. True or False: The confidence interval estimate of the population mean is constructed
around the sample mean. True
2. Estimation means assigning values to a population parameter based on the value of a
Sample Statistic
3. The confidence level is denoted by (1 )100%− .
4. The maximum error (margin of error) of the estimate for μ (based on known σ) is:
/ 2E Z
n


=
5. The parameter(s) of t distribution is (are) degrees of freedom: df
6. What criteria are required to apply the t distribution to make a confidence interval for μ?
a) The sample must be a simple random sample
b) Either the sample is from a normally distributed population or n ≥30
c) Population Standard Deviation, σ is unknown
7. Find the values for t for a t-distribution with sample size of 20 and a confidence level of
95%.
degrees of freedom: df = n – 1 = 19, 2.093t = 
8. Find the values for t for a t-distribution with sample size of 30 and a confidence level of
90%.
degrees of freedom: n – 1 = 29, 1.699t = 
9. Find the critical chi-square value for 20 degrees of freedom when the area to the left is
0.01
𝑿 𝑳
𝟐
= 𝑿 𝟎.𝟗𝟗
𝟐
= 8.260
10. Find the critical chi-square value for n=30 when the area to the right is 0.9
degrees of freedom: 29
𝑿 𝑹
𝟐
= 𝑿 𝟎.𝟗
𝟐
= 19.768
11. A statistician is interested in estimating at a 95% confidence level the mean number of
houses sold per month by all real estate agents in a large city. It is known that the
population standard deviation is 1.9.
a. How large a sample should be taken so that the estimate is within 0.65 of the
population mean?
n = [(zα/2 · σ)/E]2 = [(1.96)(1.9)/0.65]2 = 32.82 ≈ 33 (Rounded up)

2. 2
b. Find this 95% confidence interval, if the mean for such sample size is 3.4
x = 3.4, zα/2 = 1.96, E = 0.65
x – E < μ < x + E
3.4-0.65 & 3.4+ 0.65 2.75 < μ < 4.05
12. A random sample of 20 female members of a health club showed that they spend on
average 4.5 hours per week doing physical exercise with standard deviation of 0.75 hours
(assume the time spent on exercise for all female members is approximately normally
distributed) Given: n = 20, x = 4.5 hr & s = 0.75 hr
a. What is the value of the point estimate of the population mean?
point estimate of μ, the population mean is: x = 4.5 hours per week
b. Construct the 98% confidence interval for the mean time spent on physical
exercise of all such females.
degree of freedom: df = 20 – 1 = 19
CI = 0.98 → α = 0.02 → t α/2 = 2.539
E = tα/2 · (s/√ 𝒏) = (2.539)(0.75/√𝟐𝟎) = 0.4258 hours
98% CI: x – E < μ < x + E → 4.0742 < μ < 4.9258

3. 3
13. Out of 500 randomly selected adults, 300 said they were in favor of the death penalty for
a person convicted of murder. Given: n = 500, x = 300
a. What is the value of point estimate of the population proportion?
𝑝̂ =
𝑥
𝑛
=
3
5
= 0.6
b. What is the 95% confidence interval for the population proportion?
CI = 0.95 → α = 0.05 z α/2 = 1.96, 𝑞̂= 1-0.6 = 0.4
E = zα/2 √𝑝̂ 𝑞̂ /𝒏 = (1.96)√(𝟎. 𝟔)(𝟎. 𝟒)/𝟓𝟎𝟎 = 0.0429
𝑝̂ - E < p < 𝑝̂ +E 0.5571 < p < 0.6429
14. How large a sample is needed to be 99% confident that the margin of error is 0.025, for
percentage of golfers that are left-handed, if we
a. Assume that 15% of the golfers are left-handed, based on a previous study.
E = 0.025, CI = 0.99, z α/2= 2.575, 𝑝̂ = 0.15
n = (z α/2)2
𝑝̂ 𝑞̂ / E2
n = [(2.575)2
x 0.15 x 0.85] / 0.0252
n = 1352.6475; but rounds up to 1353
b. Assume that we have no prior information suggesting a possible value for the
sample proportion.

4. 4
assume 𝑝̂= 𝑞̂ = 0.5
n =
𝒛 𝜶/𝟐 (𝟎.𝟐𝟓)
𝟐
𝑬 𝟐 = [(2.575)2
(0.25)]/(0.025)2
= 2652.25 ≈ 2653 (rounded UP to 2653)
15. Given: 30, 80.5, 4.6n x s= = = (assume normal population)
a. Find 98% confidence interval for population variance
df = n – 1 = 29
49.588 = X2
R
X2
L = 14.257
b. Find 98% confidence interval for population standard deviation
square root value of part a
√
( 𝒏 − 𝟏) 𝒔 𝟐
𝑿 𝑹
𝟐 < 𝝈 < √
( 𝒏 − 𝟏) 𝒔 𝟐
𝑿 𝑳
𝟐 → √
( 𝟐𝟗)( 𝟒. 𝟔) 𝟐
𝟒𝟗. 𝟓𝟖𝟖
< 𝝈 < √
( 𝟐𝟗)( 𝟒. 𝟔) 𝟐
𝟏𝟒. 𝟐𝟓𝟕
→
𝟑. 𝟓𝟏𝟕𝟖 < 𝝈 < 𝟔. 𝟓𝟔𝟎𝟔
(𝒏−𝟏)𝒔 𝟐
𝑿 𝑹
𝟐 < 𝝈 𝟐
<
(𝒏−𝟏)𝒔 𝟐
𝑿 𝑳
𝟐 →
(𝟐𝟗)(𝟒.𝟔) 𝟐
𝟒𝟗.𝟓𝟖𝟖
< 𝝈 𝟐
<
(𝟐𝟗)(𝟒.𝟔) 𝟐
𝟏𝟒.𝟐𝟓𝟕
→
𝟏𝟐. 𝟑𝟕𝟒𝟖 < 𝝈 𝟐
< 𝟒𝟑. 𝟎𝟒