Interpreting CIs
A 90% confidence interval for the proportion of Portland residents who commute to work using public transit is (0.034, 0.058). What is the point estimate and margin of error for this CI?
P.E. = (0.034+0.058)/2 = 0.046
M.O.E = (0.058-0.034)/2 = 0.012
When do we use t* critical values?
What do we check to ensure that the sampling distribution for a sample proportion is approximately Normal? Write out the equations.
Large counts condition
n\hat{p}\geq10 and n(1-\hat{p})\geq 10
A doctor takes a sample of 200 Americans and finds that 18 have an outie belly button. Find a 99% CI for the proportion of all Americans who have an outie belly button. Assume all conditions are met.
\hat{p}+-z^{\star}sqrt((\hat{p}(1-\hat{p}))/n)
0.09+-(2.576)sqrt(((0.09)(0.91))/200)
(0.038, 0.142)
We want to find the sample size that will produce a given MOE for a proportion confidence interval. What
\hat{p}
value do we use if we do not have an estimate of the the proportion?
\hat{p}=0.5
A 90% confidence interval for the proportion of Portland residents who commute to work using public transit is (0.034, 0.058). Interpret this CI.
We are 90% confident that the proportion of all Portland residents who commute to work using public transit is captured in the interval from 0.034 to 0.058.
Find the critical value for a 96% confidence interval for a proportion where sample size n = 60
invNorm(area=0.98, mean=0, SD=1)
z* = 2.05
What do we check to ensure that the sample is a good representation of the population?
The sample is random
A school counselor wants to know the average amount of time that students spend on social media, and they want a 95% margin of error of no more than 15 minutes. A previous questionnaire about social media usage suggests that the standard deviation will be about 50 minutes. What sample size is needed?
z^{\star}\sigma/sqrt(n)=MOE
(1.96)50/sqrt(n)=15
(1.96)(50)/15=sqrt(n)
n=43 ppl
What do the values in a confidence interval represent?
Plausible values for the true parameter value.
What is the meaning of the confidence level C%?
If this procedure was repeated many times, we would expect C% of confidence intervals to capture the parameter value.
Find the critical value for a 98% confidence interval for a mean where sample size n = 45
invT(area=0.99, df=44)
t* = 2.414
table (df=40)
t* = 2.423
Why do we check the 10% condition?
To check that individual values are reasonably independent. This is necessary to calculate the standard deviation.
A dentist wants to find a 92% confidence interval for the proportion of her patients who floss every day. She wants to have a margin of error of no more than 0.1. What sample size is needed?
z^{\star}sqrt((\hat{p}(1-\hat{p}))/n)=MOE
(1.75)sqrt(((0.5)(0.5))/n)=0.1
sqrt((0.25)/n)=0.1/1.75
(0.25)/n=(0.1/1.75)^2
0.25/(0.1/1.75)^2=n
n=77
You give your boss a 95% confidence interval, and they ask for an estimate with a smaller margin of error. What are the 2 ways this can be accomplished? What are the downsides of each?
1) Decrease the confidence. Downside: we are now less confident that the interval captures the parameter.
2) Increase the sample size. Downside: this requires more money, time, and resources.
No. There are values in the CI that are below 0.5, so it is plausible that less than half of U.S. adults use an iPhone.
DAILY DOUBLE
What confidence level is needed to produce a margin of error of 0.03 if the sample size is n = 600 and the sample proportion is 0.45?
z^{\star}sqrt(((0.45)(0.55))/600)=0.03
z^{\star}=0.03/{sqrt(((0.45)(0.55))/600)}
z^{\star}=1.477
normCDF(lower=-1.477, upper=1.477, mean=0, SD=1)
Confidence level = 86%
Why do we use z* to find the sample size for a mean CI?
We cannot calculate sample size without an estimation of the population SD. If we have the population SD, we use z*.
A grad student at PSU wants to know the average number of shoes that PSU students own. They take a random sample of 50 PSU students and find that the average number of shoes is 8.4 with a standard deviation of 3.2. Find a 90% confidence interval for the mean number of shoes for all PSU students.
\bar{x}+-t^{\star}s_x/sqrt(n)
8.4+-(1.677)3.2/sqrt(50)
(7.64, 9.16)
How much larger does a sample need to be to make the margin of error half as big?
Four times larger (4n)