# Statistics Questionnaire

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ECON 15B
Formulas
1. Descriptive Statistics:
𝑥̅ =
∑𝑖 𝑥𝑖
𝑛
(∑𝑖 𝑥𝑖 )2
2
2

𝑥

(𝑥
)

𝑥̅
𝑖
𝑖
𝑖 𝑖
𝑛
𝑠2 =
=
𝑛−1
𝑛−1
𝑝̂ =
𝑝̃ =
𝑥
𝑛
𝑥+2
𝑛+4
2. Probability
𝐸(𝑥) = 𝜇 = ∑ 𝑥𝑖 × 𝑝(𝑥𝑖 )
𝑉𝑎𝑟(𝑥) = 𝜎 2 = ∑(𝑥𝑖 − 𝜇)2 × 𝑝(𝑥𝑖 )
If X has a binomial distribution with parameters 𝑛 and 𝑝, then
𝜇𝑝̂ = 𝑝
𝑝(1 − 𝑝)
𝜎𝑝̂ = √
𝑛
If 𝑥̅ is the mean of a random sample of size n from a population with mean 𝜇 and standard deviation 𝜎,
then:
𝜇𝑥̅ = 𝜇
𝜎𝑥̅ =
𝜎
√𝑛
1
ECON 15B
3. Inference
Confidence Interval: 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 ± (𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒) × (𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐)
Test Statistic:
𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐−𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐
Single-Sample
Statistic
Sample Mean
Standard Deviation
𝜎
√𝑛
Sample Proportion
𝑝(1 − 𝑝)

𝑛
Two-Sample
Statistic
Difference of Sample Means
Standard Deviation

𝜎12 𝜎22
+
𝑛1 𝑛2
Special Case when 𝜎1 = 𝜎2
1
1
√𝑠𝑝2 ( + )
𝑛1 𝑛2
(𝑛1 − 1)𝑠12 + (𝑛2 − 1)𝑠22
𝑛1 + 𝑛2 − 2
𝑡 is based on (𝑛1 + 𝑛2 − 2) 𝑑𝑓
𝑠𝑝2 =
Special Case when 𝜎1 ≠ 𝜎2
2
(
𝑣=
𝑠12 𝑠22
+ )
𝑛1 𝑛2
2
2
𝑠2
(𝑛2 )
1
+ 2
𝑛1 − 1 𝑛2 − 1
𝑡 is based on 𝑣 𝑑𝑓
Difference of Sample Proportions
𝑠2
(𝑛1 )
𝑝1 (1 − 𝑝1 ) 𝑝2 (1 − 𝑝2 )

+
𝑛1
𝑛2
Special Case when 𝑝1 = 𝑝2
1
1
√𝑝(1 − 𝑝) ( + )
𝑛1 𝑛2
(𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 − 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑)2
𝐶ℎ𝑖 − 𝑠𝑞𝑢𝑎𝑟𝑒 𝑡𝑒𝑠𝑡 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 = ∑
𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑
2
Question 4
We need to construct an 80% confidence interval for the mean commute time for all employees of this
company.
1. We will use confidence interval for the mean so we can estimate the commute time. The sample
is randomly selected so we can use assume the sample is representative from the data.
2. Constructing 80% CI.
Xbar = 27.6
SD = 5.3
N = 120
Alpha is 1 – 0.80 = 0.20
Alpha/2 = 0.10
Z = 1.28
Making the confidence interval
27.6 +/- 1.28 * 5.3 / sqrt 120
27.6 +/- 0.6193
26.9807 < mu < 28.2193 We are 80% confident that the true commute time for all employees of this company is between 26.9807 and 28.2193. Question 5 1. We will use confidence interval for proportion since we have a sample proportion and we can use normal approximation because n*p > 5 and n*(1-p) > 5
2. 99% confidence interval calculations.
P = 0.55
N = 647
Z = 2.57 for 99% CI (using table)
0.55 ∗ (1 − 0.55)
0.55 ± (2.57 ∗ √
)
647
0.55 +/- 0.0503
0.4997 < P < 0.6003 3. interpretation We are 99% confident that the true proportion of parents in US with children under 12 saying they would get their child vaccinated against covid if such a vaccine were available is between 49.97% and 60.03% Question 6. For this case we need to use this formula 𝑍 ∗ 𝑠𝑖𝑔𝑚𝑎 2 𝑛=( ) 𝑒𝑟𝑟𝑜𝑟 Error is 10, the standard deviation is 50, and the Z value for 95% confident is 1.96 1.96 ∗ 50 2 𝑛=( ) = 96.04 = 97 10 So the sample size that we need to take if the group wants to be 95% confident is 97. Question 7. 1. That is wrong, remember that the confidence interval is not a probability, its about confident. 2. This is wrong, the confident is about population mean, not sample mean. 3. That is correct! 4. That is correct! 5. Wrong, we are 99% confident, we are not talking about percentage of students when we are constructing the confidence interval. 6. Wrong, again we are talking about confidence interval, not percentage of people in the sample. Question 1. The random sample is size 3 so we will take 3 different numbers from the 2 that we have from the population. We will find the sample mean of our sample. And we will find the probability by multiplying the probability of the numbers we are using in a specific sample. sample 3,3,3 3,3,0 3,0,3 0,3,3 3,0,0 0,3,0 0,0,3 0,0,0 sample mean probability 3 0.027 2 0.063 2 0.063 2 0.063 1 0.147 1 0.147 1 0.147 0 0.343 sum 1 Properties of sampling distribution The sample mean will be the same value as population mean The standard deviation of the sampling distribution would be standard deviation of population divided by square root of the sample size. The sampling distribution would be normally distributed if the sample size is greater than 30, no matter the distribution of the population. 𝑠𝑎𝑚𝑝𝑙𝑒 𝑚𝑒𝑎𝑛 = ∑ 𝑥 ∗ 𝑃(𝑋) Sample mean = = 3 ∗ 0.027 + 2 ∗ 0.063 + 2 ∗ 0.063 + 2 ∗ 0.063 + 1 ∗ 0.147 + 1 ∗ 0.147 + 1 ∗ 0.147 + 0 ∗ 0.343 Standard deviation would be… 𝑆𝐷 = √(∑ 𝑥 2 ∗ 𝑃(𝑋)) − 𝑚𝑒𝑎𝑛2 SD = √(32 ∗ 0.027 + 22 ∗ 0.063 + 22 ∗ 0.063 + 22 ∗ 0.063 + 12 ∗ 0.147 + 12 ∗ 0.147 + 12 ∗ 0.147 + 0^2 ∗ 0.343) − 𝑚𝑒𝑎𝑛2 Question 2. We are going to construct the 95% confidence interval for mean using T distribution because the sample size is small. Xbar = 30 SD = 6 N=8 T = 2.36 by looking T table The distribution is normal so we can say that the conditions are valid. Constructing 95% CI… 30 +/- (2.36 * 6 / sqrt 8) 30 +/- 5.0063 24.9937 < mu < 35.0063 We are 95% confident that the true mean weight of backpacks is between 24.9937 and 35.0063 pounds. Question 3. P( Xbar < 2.5) Calculating Z first. Z = (2.5 – 2.75) / (1.25 / sqrt 49) Z = -1.40 Now using Z table we will get the probability… Probability is 0.0808 Part 2. We know that 67% of sample means are greater than 2.64 P(X > 2.64) = 0.67
We can find Z value with that information.
Z = -0.44 (it is negative because the area below 2.64 is lower than 0.50)
Now using the formula for Z value, we will get sigma
Z = (x – mu) / (SD / sqrt N)
SD = (x – mu) * sqrt N / Z
SD = (2.64 – 2.75) * sqrt 49 / -0.44
SD = 0.0357
So population standard deviation is 0.0357 for the conditions of part 2 problem
Question 6
Private colleges and universities rely on money contributed by individuals and corporations for their
operating expenses. Much of this money is invested in a fund called an endowment, and the college
spends only the interest earned by the fund. A recent survey of eight private colleges in the United
States revealed the following endowments (in millions of dollars): 60.2, 47.0, 235.1, 490.0, 122.6,
177.5, 95.4, and 220.0. Summary statistics yield and Calculate a 90% confidence interval for the
mean endowment of all private colleges in the United States.
Part a: State the assumptions that are required for this interval to be valid.
Part b: Construct a 90% confidence interval.
Part c: Interpret the confidence interval in (b).
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Question 7
A Florida newspaper reported on the topics that teenagers most want to discuss with their parents.
The findings, the results of a poll, showed that 46% would like more discussion about the family’s
financial situation, 37% would like to talk about school, and 30% would like to talk about religion.
These and other percentages were based on a national sampling of 532 teenagers. Estimate the
proportion of all teenagers who want more family discussions about school. Use a 95% confidence
level.
Part a: State the assumptions that are required for this interval to be valid.
Part b: Construct a 95% confidence interval.
Part c: Interpret the confidence interval in (b).
Question 8
An auditor checks a sample of 225 randomly chosen transactions from among the thousands
processed in an office. Thirty-five contain errors in crediting or debiting the appropriate account.
Part a: Does this situation meet the conditions required for a confidence interval for the population
proportion?
Part b: Find the 95% confidence interval for the proportion of all transactions processed in this office
that have these small errors.
Part c: Interpret the confidence interval.
Part d: Managers claim that the proportion of error is about 10%. Does that seem reasonable?
Question 9
One of the characteristics that determine success of credit cards is the average outstanding balance.
Prior to extending an offer, the credit company wants to estimate the average outstanding balance.
The company randomly selects 140 customers. The average monthly balance is calculated as
\$1,990.50 with a standard deviation of \$2,833.33.
Part a: Construct a 95% confidence interval estimate.
Part b: Interpret the confidence interval.
Part c: State the conditions required for this interval to be valid.
Question 1
Compute the sampling distribution for two tosses of a fair coin. Assume x=1 for heads and x=0 for
tails.
Question 2
According to a 1995 study, the mean family income in the US was \$38,000 with a standard
deviation of 21,000. If a consulting agency surveys 49 families at random, what is the probability
that it finds a mean family income of more than \$41,500?
Question 3
One year, the distribution of salaries for professional sports players had mean \$1.5 million and
standard deviation \$0.9 million. Suppose a sample of 400 major league players was taken. Find the
approximate probability that the average salary of the 400 players that year exceeded \$1.1 million.
Question 4
How much money does the average professional football fan spend on food at a single football
game? That question was posed to 45 randomly selected football fans. The sample results provided a
sample mean and standard deviation of \$18.00 and \$3.15, respectively.
Part a: State the assumptions that are required for this interval to be valid.
Part b: Construct a 99% confidence interval.
Part c: Interpret the confidence interval in (b).
Question 5
The average cost per night of a hotel room in New York City is \$273. Assume this estimate is based
on a random sample of 45 hotels and that the sample standard deviation is \$65.
Part a: Construct the 92% confidence interval estimate of the population mean.
Part b: Interpret the confidence interval.
Part c: What are the assumptions required for this confidence interval to be valid?
Part d: What will happen to the width of the confidence interval if the confidence coefficient would
increase to .95?
Part e: What would happen to the width of the confidence interval if the sample had 100 hotels?

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