Chapters 19 and 23 Notes: One-Sample z-Interval and One-Sample t-Interval
A. Confidence Interval: An interval that covers the true parameter (proportion or mean), given our best estimate of where the parameter is and how certain (confidence level) we are that it’s within some range.
1. Based on the sample proportion (^p) or sample mean (ŷ)
2. Based on the assumption that the sampling distribution model will follow the shape of a Normal model.
3. Must follow the same Assumptions/Conditions: a. Independence a. Randomization b. 10% Assumption c. Success/Failure Condition: np ≥ 10 and nq ≥ 10 (for proportions)
4. Confidence Level (z* or t*): A numerical value assigned to determine how certain the confidence interval is:
5. Example (proportion): During the 2004 presidential election, the Gallup organization asked registered voters if there was “Some chance they could vote for the other candidate” besides their expressed first choice. Out of the 909 sampled voters, only
18% indicated that there was some chance they might switch. The resulting 95% confidence interval is 0.18 ± 0.025 = 15.5% to 20.5%: a. What does this interval mean (in context)?
We are 95% confident that that the true proportion of registered voters in 2004 who would consider voting for the other candidate is between 15.5% and 20.5%. b. What does the confidence mean in context?
About 95% of all random samples of 909 registered voters will produce the true proportion of voters who would consider voting for the other candidate.
6. Example (mean): During 2008, residents were concerned that vehicles traveling on
Triphammer Road often exceed the posted speed limit of 30 miles per hour. The local police placed a radar speed detector to determine the mean speed of all vehicles traveling on the road. Out of a sample of 34 cars, it was determined that the mean speed of the cars was 31.0 mph with a standard deviation of 4.24 mph. The resulting
90% confidence interval for the mean speed was 31.0 ± 1.5 = 29.5 – 32.5 mph. a. What does this interval mean in context?
We are 90% confident that true mean speed of vehicles on Triphammer Road in 2008 is between 29.5 and 32.5 mph. b. What does the confidence interval mean in context?
About 90% of all samples of 34 vehicles traveling down Triphammer Road will produce the true mean speed of vehicles on the road.
B. One-Proportion z-interval: The approximate confidence interval calculated and interpreted based on the sample proportion (^p), the confidence level (z*), and the
Margin of Error (ME).
1. One-Proportion z-interva l: ^p ± z* ∙ SE(^p), (ME = z* ∙ SE(^p))
2. SE(^p) = √((^p^q)/n)
3. Confidence Level (z* or t*): A numerical value assigned to determine how certain the confidence interval is:

a. 80% = 1.282, 90% = 1.645, 95% = 1.960, 98% = 2.326, 99% = 2.576 b. Most common confidence values: 90% = 1.645 and 95% = 1.960
4. Example: In May 2002, the Gallup Poll asked 537 randomly sampled adults the question “Generally speaking, do you believe the death penalty is applied fairly or unfairly in this country today?” Of these, 53% answered “Fairly”. With 95% confidence, what can we conclude from this survey?
Adult responses to the question are independent.
The 537 adults were selected at random.
The 537 adults represent less than 10% of all adults in the U.S.
Success/Fail ure: 0.53(537) = ~285 ≥ 10, 0.47(537) = ~252 ≥ 10, the sample size is large enough.
The conditions for a Normal model are satisfied
One-Proportion z-interval: n = 537, ^p = 0.53, SE = √((0.53 ∙ 0.47)/537) = 0.022, C = 95%, z* = 1.960
0.53 ± 1.960(0.022) = 0.53 ± 0.043 or (0.487, 0.573)
Conclusion: I am 95% confident that the true proportion of adults who think the death penalty is applied fairly is between 48.7% and 57.3%.
What does your confidence mean?
About 95% of all random samples of 537 U.S. adults will produce the true proportion of people who think that the death penalty is applied fairly.
C. One Sample t-distribution: t-distributions are unimodel, symmetric, and bell-shaped
(approximately Normal).
1. Degrees of Freedom (t* = n - 1), Appendix G (p. A-72)
-as the t* increases, the t-distribution will approach Normal (actual confidence value, see B.3.a)
-t* depends upon the particular confidence level used during the test
C = 95%, n = 30, t*(30 - 1) = t*29 = 2.045
C = 90%, n = 30, t*(30 – 1) = t*29 = 1.699
2. One Sample t-interval: a. ŷ ± t*
(n
–
1)
SE(ŷ), SE(ŷ) = sd/√n, ŷ = population mean
sd = standard deviation
3. Example: Hoping to lure more shoppers downtown, a city builds a new public parking garage in the central business district. The city plans to pay for the structure through parking fees. During a two-month period (44 weekdays), daily fees collected averaged
$126 with a standard deviation of $15.
N($126,$15), n = 44 weekdays a. What conditions must you make in order to use these statistics for inference?
Daily fees from one weekday are independent from all other weekdays. The 44 weekdays are a representative sample of all possible weekdays. The 44 weekdays represent less than 10% of all possible weekdays. Since the sample was 44 weekdays, the sample size is large enough.
*Nearly Normal Condition (for means only): The distribution of weekdays is assumed to be unimodel and symmetric (Normal) (You must check this condition by making a histogram, if you can. You must be given the distribution. If the distribution is not given, then we can only make the assumption that the distribution is Normal).

b. Write a 90% confidence interval for the mean daily income this parking garage will generate.
C = 90%, df = 44-1 = 43, t* = 1.684
126 ± 1.684(15/√44)
126 ± 3.81
($122.19, $129.81) c. Explain in context what this confidence interval means.
We are 90% confident that the true mean of daily parking fees is between $122.19 and
$129.81. d. Explain what “90% confidence” means in this context.
90% of all random samples weekdays of size 44 would produce the true mean of daily parking fees. e. The consultant who advised the city on this project predicted that parking revenues would average $130 per day. Based on your confidence interval, do you think the consultant could have been correct? Explain.
The consultant may not be correct. Since $130 falls just above the confidence interval, there is some evidence to suggest that the average parking fees would be less than
$130.