Download Lesson 1 - Homework

Lesson 7 – Practice Problems and Homework 7
Notes:
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You may use minitab whenever appropriate and include the minitab output in your homework. To
place Minitab output into your homework, highlight the portion of the Minitab printout you
want, copy that to the clipboard, and then paste it into the document.
You need to check necessary conditions, e.g. sample size larger than 30 or normality, before
proceeding to compute the t-intervals. When sample size is larger than 30, one does not need to
check for Normality since population does not need to have Normal distribution and yet sample
mean (do not confuse it with the population) has a Normal distribution by Central Limit Theorem.
One only needs to use boxplot to check for extreme outliers. (One can use t-procedure to estimate
population mean when sample size is large and there is no extreme outliers.)
I. Practice problem and its solution (no need to submit problems in this section):
[The homework problems that need to be submitted are given in section II after section I]
Practice problem: Consumer reports tested 15 brands of vanilla yogurt and found the following numbers
of calories per serving: 160, 200, 220, 230, 120, 180, 140, 130, 170, 180, 80, 120, 100, 170, 190. Place a
90% confidence interval on the average number of calories per serving for vanilla yogurt. Use Minitab to
find the interval. Do not hand-compute.
Solution:
a. The sample size is small n=15(less than 30). So first we need to use the normal probability plot in
Minitab to check if this sample comes from a normal distribution.
Minitab Commands for drawing the normal probability plot:
Minitab>Graph>Probability Plot
Output is:
Probability Plot of calories
Normal - 95% CI
99
Mean
StDev
N
AD
P-Value
95
90
159.3
43.50
15
0.184
0.892
Percent
80
70
60
50
40
30
20
10
5
1
0
50
100
150
calories
200
250
300
Conclusion: The normal probability plot above suggests that the data are normally distributed since the data
all fall within the band.
b. The 90% t confidence intervals can be used here because the data is normal distributed.
Minitab Commands for finding the one-sample t-interval:
Minitab>Stat>Basic Stat>1-Sample t
Get into Options and change the confidence level to the needed one:
(Note that for now you can ignore the test mean part since that deals with testing.)
Minitab will give you the following output:
One-Sample T: calories
Variable
calories
N
15
Mean
159.333
StDev
43.502
SE Mean
11.232
90% CI
(139.550, 179.116)
Conclusion: we are 90% confident that the average number of calories per serving for vanilla yogurt is
between 139.55 and 179.116 calories.
II. Homework problems (submit to Dropbox for HW 7 by its due day)
1. The ability to read rapidly and simultaneously maintain a high level of comprehension is often a
determining factor in the academic success of many high school students. A school district is considering a
supplemental reading program for incoming freshmen. Prior to implementing the program, the school runs
a pilot program on a random sample of n = 20 students. The students were thoroughly tested to determine
reading speed and reading comprehension. Based on a fixed-length standardized test reading passage, the
following reading times (in minutes) and increases in comprehension scores (based on a 100-point scale)
were recorded. The data are given below.
Student 1 2
3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 n y
s
Reading
Time
5 7 15 10 8 7 10 11 9 13 10 8 11 8 10 8
7
6 11 8 20 9.10 2.404
Comprehension 66 76 76 90 81 75 95 98 88 73 90 70 91 83 100 85 76 69 91 95 20 83.40 10.32
a Place a 95% confidence interval on the mean reading time for all incoming freshmen in the district. Use
minitab to find the interval, do not hand compute.
b. Plot the reading time using a normal probability plot.
c. Provide an interpretation of the interval estimate in part a.
d. How would your inference change by using a 97% confidence interval in place of the 95% confidence
interval?
2. Hand compute to find the answer to part a of the previous problem 1.
3. An office manager has implemented an incentive plan that she thinks will reduce the mean time required
to handle a customer complaint. The mean time for handling a complaint was 30 minutes prior to
implementing the incentive plan. After the plan was in place for several months, a random sample of the
records of 34 customers who had complaints revealed a mean time of 27.6 minutes with a standard
deviation of 3.7 minutes.
a. Give a point estimate of the mean time required to handle a customer complaint.
b.What is the standard deviation of the point estimate given in a?
c. Construct a 95% confidence on the mean time to handle a complaint after implementing the plan.
Interpret the confidence interval for the office manager.
[Hint: use minitab to find the t-value for 33 degrees of freedom, point estimate refers to sample mean,
standard deviation of the point estimate refers to the standard deviation of sample mean which is also called
standard error.]
4. Answer “true” or “false” for each question.
a. If a larger number of random samples are selected and we form the 95% confidence interval for
each sample mean, the population mean will lie in about 95% of these confidence intervals.
b. The 95% confidence interval around a given sample mean is wider than the 90% confidence
interval around that mean.
c. If sample size is larger than 30, there is a 95% chance that the population mean will be in the
confidence interval.
d. If a very large number of random samples are selected, there is a 95% chance that one of the
sample means is equal to the population mean.
5. A test was conducted to determine the length of time required for a student to read a specified amount
of material. The students were instructed to read at the maximum speed at which they could still
comprehend the material because a comprehension test would be given on the material. Sixteen thirdgrade students were randomly selected from a large school district, and the results of their test are as
follows (in minutes):
25 18 27 29 20 19 16 24 37 21 24 19 20 28 31 22
a. Estimate the mean length of time required for all third-grade students to read the material, using a
99% confidence interval.
b. What is the population for which the confidence interval in part (a) is applicable?
c. Provide an interpretation of the confidence interval computed in part (a).
6. City housing department wants to estimate the average rent for rent-controlled apartments. They need to
determine the number of renters to include in the survey in order to estimate the average rent to within $97
using a 92% confidence interval. From past results, the rent for controlled apartments ranged from $450 to
$1500 per month. How many renters are needed in the survey? (Need to only use the crude method.)
7. If the level of confidence in above question is increased to 98%, what sample size is needed? (Need to
only use the crude method.)