Formula/ Equation
Finding Required Sample Size
Proportion
Finding Confidence Interval
Vocabulary
100
Variable for Critical Value for finding the sample size
What is Z*?
100
SD=10, Confidence level 90%, Margin of Error 16
Smallest required sample size=0
100
p=p'±[Z(a/2)√p'(1-p’)/n
P' = 0.08+.16/2 = .12
[Z√(0.12(1-0.12))/150]=0.04 therefore Z=1.51
p(Z=1.51)=0.4345
p(Z=1.51)=0.4345,
p(Z)⋅2=0.8690
p(Z)⋅2=0.8690 or 86.90%
What is The Dundee Dog Training School has a larger than average proportion of clients who compete in competitive professional events. A confidence interval for the population proportion of dogs that compete in professional events from 150 different training schools is constructed. The lower limit is determined to be 0.08 and the upper limit is determined to be 0.16. Determine the level of confidence used to construct the interval of the population proportion of dogs that compete in professional events.
100
You want to rent an unfurnished one-bedroom apartment in Durham, NC next year. The mean monthly rent for a random sample of 60 apartments advertised on Craig’s List (a website that lists apartments for rent) is $1000. Assume a population standard deviation of $200. Construct a 95% confidence interval.
($1000 – 1.96*$200/sqrt(60), $1000 + 1.96*$200/sqrt(60))
($949.39, $1050.61)
We are 95% confident that the interval ($949.39, $1050.61)
covers the true mean monthly rent of Durham apartments listed on
Craig’s List.
100
To say that we are 95% confident is shorthand for "95% of possible samples of a given size from this population will result in an interval that captures the unknown parameter."
What is a confidence level?
200
Margin of error definition
What is an amount (usually small) that is allowed for in case of miscalculation or change of circumstances?
200
P-hat, .50, 95% confidence level, margin of error 4
N=(1960/.04)^2 (.5) (1-.5)=600.25
N Required= 601
200
x=300 and n=500
p’=x/n=300/500=.600
q’=1-p’=1-.600=.400
Since confidence level = .90, then a=1 -confidence level = (1-.90)=.10(a/2)=.05
Z(a/2=Z.05=1.645
p’-Za√p’q’/n ≤p≤p’+Za√p’q’/n
0.564≤p≤.636
Ninety percent of all confidence intervals constructed in this way contain the true values for the population percent of accounts receivable that are overdue 30 days.
What is A financial officer for a company wants to estimate the percent of accounts receivable that are more than 30 days overdue. He surveys 500 accounts and finds that 300 are more than 30 days overdue. Compute a 90% confidence interval for the true percent of accounts receivable that are more than 30 days overdue, and interpret the confidence interval proportion.
200
How large a sample of one-bedroom apartments above would be needed to estimate the population mean within plus or minus $50 with 90% confidence?
The z multiplier for a 90% confidence interval is 1.645. We want z*σ/sqrt(n) to equal $50.
$𝟓𝟎=𝒛∗ 𝝈 =𝟏.𝟔𝟒𝟓∗𝟐𝟎𝟎=𝟑𝟐𝟗 𝒏𝒏𝒏
Solve for n. sqrt(n) = 329/50 = 6.58. Square both sides and you get an n equal to 43.3. To calculate a margin of error +/- $50, you would need to randomly sample 44 rental apartments on Craig’s List.
200
"We are C% confident that the interval from ___ to ___ captures the actual value of the [population parameter in context]."
estimate plus/minus margin of error
What is a confidence interval?
300
When the standard deviation is unknown
What is x̄ ± t* σ / (√n)
300
p̂=0.5, Confidence level 95%, Margin of error .03
Smallest required sample size is 1069
300
Let X = the number of people in the sample who have cell phones. X is binomial.
𝑋~𝐵(500,421/500)
n=500
x=421
p’=421/500=.842
We estimate with 95 percent confidence that between 81 percent and 87.4 percent of all adult residents of this city have cell phones.
What is Suppose that a market research firm is hired to estimate the percentage of adults living in a large city who have cell phones. Five hundred randomly selected adult residents in this city are surveyed to determine whether they have cell phones. Of the 500 people surveyed, 421 responded yes, they own cell phones. Using a 95 percent confidence level, compute a confidence interval estimate for the true proportion of adult residents of this city who have cell phones.
300
Duncan Jones kept careful records of the fuel efficiency of his car. After the first 100 times he filled up the tank, he found the mean was 23.4 miles per gallon (mpg) with a population standard deviation of 0.9
mpg. Compute the 95 percent confidence interval for his mpg.
(23.4 – 1.96*0.9/sqrt(100), 23.4 +1.96*0.9/sqrt(100))
We are 95% confident that the interval (23.22 mpg, 23.58 mpg) includes the true mean mpg of Duncan’s automobile.
300
Depends on both the confidence level C and the sampling distribution of the statistic.
What is critical value?
400
The formula for Confidence interval for sigma known means
What is x̄ ± z* σ / (√n)?
400
A university wants to know more about the knowledge of students regarding international events. The are concerned that their students are uninformed in regards to new from other countries. A standardized test is used to assess students knowledge of world events (national reported mean=65, S=5). A sample of 30 students are tested (sample mean=58, Standard Error=3.2). Compute a 99 percent confidence interval based on this sample's data. How do these students compare to the national sample?
What is the 90% confidence interval for BMI? (Note that Z=1.645 to reflect the 90% confidence level.)
400
Random: SRS or Random experiment
Normal: if the population distribution is normal the sample distribution is too, CLT for x
Independent: When sampling without replacement, the 10% condition.
What are the Conditions for Constructing a Confidence Interval?
500
The formula of Confidence Interval of proportions
What is 
± z* √(1-) / N
500
State: What parameter do you want to estimate, and at what confidence level?
Plan: Identify the appropriate inference method. Check conditions.
Do: If the conditions are met, perform calculations.
Conclude: Interpret your interval in the context of the problem.
What is the four step process?