T Objectives: Confidence interval and significance tests Author: Will A. Eagon

STAT 350 (Fall 2014)
Lab 6: Minitab Solutions
1
Author: Will A. Eagon
Lab 6 - One Sample T Confidence Interval and Test
Objectives: Confidence interval and significance tests
A. (30 points) Number of Friends on Facebook (Data Set: FACEBOOKFRIENDS) Facebook
provides a variety of statistics on their Web site that detail the growth and popularity of the site. One
such statistic is that the average user has 130 friends. Consider the following SRS of n = 30 Facebook
users from a large university.
Solution:
File → Open Worksheet → Files of type: Text (*.txt) → facebookfriends.txt → Open
1.
(5 pts) Do you think these data are Normally distributed? Use graphical methods to
examine the distribution. Write a short summary of your findings.
Graph->Histogram->Simple->Select Variables: C1-> Data view->Select normal (let Minitab estimate
parameters) and let Smoother select Lowess with the default values->OK->OK.
Below the histogram with the normal curve superimposed and Lowess curve as the kernel illustrate
that is dataset may not be normally distributed.
STAT 350 (Fall 2014)
Lab 6: Minitab Solutions
2
Author: Will A. Eagon
Graph  Probability Plot  Single
Graph variables: C1
Distribution  Data Display  Be sure that ‘Show confidence interval’ is NOT checked. → OK
Scale  Y-Scale Type  Score (You will be marked off if the default selection is used) → OK → OK
The data are roughly normal, since all the points in the QQplot are closed to the straight line and the
empirical density curve is closed to the theoretical normal density curve.
2.
(5 pts) Is it appropriate to use the t methods of this section to compute a 95%
confidence interval for the mean number of Facebook users at this large
university? Explain why or why not.
Solution:
Since the data are roughly normal, it is appropriate to use the t methods for the mean number of
Facebook users at this large university.
3.
(5 pts) Find the mean, standard deviation, standard error, and margin of error for 95%
confidence. Compute the 95% CI for μ using these numbers. If the numbers are
obtained via the package, please include the appropriate output. If the numbers are
calculated by hand, please show your work.
Solution:
Stat → Basic Statistic → Display descriptive statistic → “Variables”: Friends → OK
Descriptive Statistics: Friends
Variable
Friends
N
30
N*
0
Mean
119.07
SE Mean
5.40
StDev
29.57
Minimum
72.00
Q1
98.50
Median
115.00
Q3
139.75
Maximum
193.00
Mean: 119.07, standard deviation: 29.57, standard error = 5.40
margin of error = (t*(29)) (standard error) = (2.045) (5.40) = 11.043
CI: x̄  m = 119.07  11.043 ==> (108.027, 130.113)
4.
(5 pts) Report the 95% confidence interval for μ, the average number of friends for
Facebook users at this large university. This answer is obtained from the software
package so the output needs to be reported. Compare with your answer in part 3.
Solution:
Stat → Basic Statistics → 1-sample t → “One or more samples, each in a column”:Friends → Options → Confidence level: 95.0 → Alternative hypothesis: mean ≠ hypothesized mean → OK → OK
STAT 350 (Fall 2014)
Lab 6: Minitab Solutions
3
Author: Will A. Eagon
One-Sample T: Friends
Variable
Friends
N
30
Mean
119.07
StDev
29.57
SE Mean
5.40
95% CI
(108.03, 130.11)
The results are the same.
5.
(5 pts) Interpret your 95% confidence interval for μ obtained in previous question.
Solution:
We are 95% confidence that the population average number of friends for Facebook users at this large
university is between 108.03 and 130.11.
B (50 points) Counts of Picks in a 1-lb bag (Data Set: PICKCOUNT -website) A guitar supply
company must maintain strict oversight on the number of picks they package for sale to
customers. Their current advertisement specifies between 900 and 1000 picks in every bag. An SRS of
thirty-six 1-pound bags of picks were collected as part of a Six Sigma Quality Improvement effort
within the company. The number of picks in each bag are shown in the following table.
Solution:
File → Open Worksheet → Files of type: Text (*.txt) → pickcount.txt → Open
1.
(12 pts) Create a histogram, boxplot, and a Normal quantile plot of these counts.
Solution:
Graph->Histogram->Simple->Select Variables: C1-> Data view->Select normal (let Minitab estimate
parameters) and let Smoother select Lowess with the default values->OK->OK.
STAT 350 (Fall 2014)
Lab 6: Minitab Solutions
4
Author: Will A. Eagon
Graph->Boxplot->Simple->Select C1 and data view mean symbol->OK
Graph->Probability Plot->Simple->Data Display (Uncheck the confidence interval)->OK->Ok
2 (4 pts) Write a description of the distribution. Comment on the skewness and Normality of the
data. Note if there are any outliers.
Solution:
From the graphical summary above, we can find that the data is not skewed and is normally
distributed. This is no outliers
3. (5 pts) Based on your observations in part (1), is it appropriate to analyze these data using
the t procedures? Briefly explain your response.
Solution:
It is appropriate to analyze these data using the t procedure since the data is symmetric and roughly
normal without any outlier.
4. (3 pts) Find the mean, the standard deviation, and the standard error of the mean for this sample.
Solution:
Stat->Basic Statistics->1-sample t->Insert column name in “One or more samples, each in a
column”->Check Perform Hypothesis, Set Hypothesize mean to 925->Optons (Mean>Hypothesize
mean)->Confidence level: 95.0->OK->OK
STAT 350 (Fall 2014)
Lab 6: Minitab Solutions
5
Author: Will A. Eagon
One-Sample T: PickCount
Test of μ = 925 vs > 925
Variable
PickCount
N
36
Mean
938.22
StDev
24.30
SE Mean
4.05
95% Lower Bound
931.38
T
3.27
P
0.001
The results are highlighted in yellow.
Mean: 938.22, standard deviation: 24.30, standard error: 4.05.
5. (5 pts) Find the 95% lower confidence bound for the mean number of picks in a 1-pound bag.
Solution:
The results are highlighted in green above.
The lower bound is 931.38
6. (8 pts) Do these data provide evidence that the average number of picks in a 1-pound bag is
greater than 925? Carry out a test of significance using the four-step procedure, with a
significance level of 5%, state your hypotheses, the value of test statistic, the P-value, and your
conclusions. Please provide the relevant output required for the steps and include all four steps
written out by hand.
Solution:
The results are highlighted in blue above.
Step 0: Definition of the terms
 is the population average number of picks in a 1-pound bag.
Step 1: State the hypotheses
H0:  = 925
Ha:  > 925
Step 2: Find the Test Statistic, report DF.
tt = 3.27
DF = 36 – 1 = 35
Step 3: Find the p-value:
P-value = 0.001
Step 4: Conclusion:
 = 1 – C = 1 – 0.95 = 0.05
Since 0.001 < 0.05 , we should reject H0
The data provides enough evidence (P-value = 0.001) to the claim that the average number of picks
in a 1-pound bag is greater than 925
7. (8 pts) Do these data provide evidence that the average number of picks in a 1-pound bag is
greater than 935? Carry out a test of significance using the four-step procedure, with a significance
level of 5%. Please see the directions for Part 6.
Solution:
Stat->Basic Statistics->1-sample t->Insert column name in “One or more samples, each in a
column”->Check Perform Hypothesis, Set Hypothesize mean to 935->Options (Mean>Hypothesize
mean)->Confidence level: 95.0->OK->OK
STAT 350 (Fall 2014)
Lab 6: Minitab Solutions
6
Author: Will A. Eagon
One-Sample T: PickCount
Test of μ = 935 vs > 935
Variable
PickCount
N
36
Mean
938.22
StDev
24.30
SE Mean
4.05
95% Lower Bound
931.38
T
0.80
P
0.216
Step 0: Definition of the terms
 is the population average number of picks in a 1-pound bag.
Step 1: State the hypotheses
H0:  = 935
Ha:  > 935
Step 2: Find the Test Statistic, report DF.
tt = 0.80
DF = 36 – 1 = 35
Step 3: Find the p-value:
P-value = 0.216
Step 4: Conclusion:
 = 1 – C = 1 – 0.95 = 0.05
Since 0.216 > 0.05 , we fail to reject H0
The data does not provide enough evidence (P-value = 0.216) to the claim that the average number
of picks in a 1-pound bag is greater than 935
8. (5 pts) Compare your conclusions for parts (5), (6) and (7). Are they the same or different?
Solution:
In part 5, we find the lower bound is around 931. In part 6, we find that the mean is greater than
925. Thus 5 and 6 are consistent because if  is greater than 931, it is greater than 925. However, in
7, we find that the mean is not greater than 935. This is also consistent with part 5 because if  is
greater than 931, it is not necessarily greater than 935.
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