.

ECON 15B

Formulas

1. Descriptive Statistics:

π₯Μ
=

βπ π₯π

π

(βπ π₯π )2

2

2

β

π₯

β

β

(π₯

)

β

π₯Μ

π

π

π π

π

π 2 =

=

πβ1

πβ1

πΜ =

πΜ =

π₯

π

π₯+2

π+4

2. Probability

πΈ(π₯) = π = β π₯π Γ π(π₯π )

πππ(π₯) = π 2 = β(π₯π β π)2 Γ π(π₯π )

If X has a binomial distribution with parameters π and π, then

ππΜ = π

π(1 β π)

ππΜ = β

π

If π₯Μ
is the mean of a random sample of size n from a population with mean π and standard deviation π,

then:

ππ₯Μ
= π

ππ₯Μ
=

π

βπ

1

ECON 15B

3. Inference

Confidence Interval: π π‘ππ‘ππ π‘ππ Β± (ππππ‘ππππ π£πππ’π) Γ (π π‘ππππππ πππ£πππ‘πππ ππ π π‘ππ‘ππ π‘ππ)

Test Statistic:

π π‘ππ‘ππ π‘ππβπππππππ‘ππ

π π‘ππππππ πππ£πππ‘πππ ππ π π‘ππ‘ππ π‘ππ

Single-Sample

Statistic

Sample Mean

Standard Deviation

π

βπ

Sample Proportion

π(1 β π)

β

π

Two-Sample

Statistic

Difference of Sample Means

Standard Deviation

β

π12 π22

+

π1 π2

Special Case when π1 = π2

1

1

βπ π2 ( + )

π1 π2

(π1 β 1)π 12 + (π2 β 1)π 22

π1 + π2 β 2

π‘ is based on (π1 + π2 β 2) ππ

π π2 =

Special Case when π1 β π2

2

(

π£=

π 12 π 22

+ )

π1 π2

2

2

π 2

(π2 )

1

+ 2

π1 β 1 π2 β 1

π‘ is based on π£ ππ

Difference of Sample Proportions

π 2

(π1 )

π1 (1 β π1 ) π2 (1 β π2 )

β

+

π1

π2

Special Case when π1 = π2

1

1

βπ(1 β π) ( + )

π1 π2

(πππ πππ£ππ β ππ₯ππππ‘ππ)2

πΆβπ β π ππ’πππ π‘ππ π‘ π π‘ππ‘ππ π‘ππ = β

ππ₯ππππ‘ππ

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 β 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 β (1 β 0.55)

0.55 Β± (2.57 β β

)

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
π β π ππππ 2
π=(
)
πππππ
Error is 10, the standard deviation is 50, and the Z value for 95% confident is 1.96
1.96 β 50 2
π=(
) = 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.
π πππππ ππππ = β π₯ β π(π)
Sample mean = = 3 β 0.027 + 2 β 0.063 + 2 β 0.063 + 2 β 0.063 + 1 β 0.147 + 1 β 0.147 +
1 β 0.147 + 0 β 0.343
Standard deviation would beβ¦
ππ· = β(β π₯ 2 β π(π)) β ππππ2
SD =
β(32 β 0.027 + 22 β 0.063 + 22 β 0.063 + 22 β 0.063 + 12 β 0.147 + 12 β 0.147 + 12 β 0.147 + 0^2 β 0.343) β ππππ2
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% CIβ¦
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 β 2.75) / (1.25 / sqrt 49)
Z = -1.40
Now using Z table we will get the probabilityβ¦
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 β mu) / (SD / sqrt N)

SD = (x β mu) * sqrt N / Z

SD = (2.64 β 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).

>

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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