test scores
69
57
55
63
48
56
59
58.1428571
13.8868/6=2.3145
(10 points) A firm that installs large computers has introduced a new training course for installers. The first class of seven has just
completed the course. Their scores are 69, 57, 55, 63, 48, 56, and 59. Assuming that these seven people are a random sample of
trainees and the scores are normally distributed, find a 95% confidence interval estimate of the mean score for the new training
course.
α = .05
n=7
standard deviation = 1.5213
( 56.73 ≤ µ ≤ 59.55)
population mean
population std dev
140
30
sample mean =
sample size =
a=
136
50
0.1
H0: µ = 140
HA: µ ≠ 140
z = (136- 140) / (30/ sqrt 50)
-4/4.2426
z=-.9428
α= .1
.1736 ≥ .1 = DNR
ATMs must be stocked with enough cash to satisfy customers making withdrawals over an entire weekend. However, if too much cash is
unnecessarily kept in the ATMs, the bank is foregoing the opportunity of investing the money and earning interest. Suppose that at a
particular branch the population mean amount of money withdrawn from ATMs per customer transaction over the weekend has
historically been $140 with a population standard deviation of $30. The branch manager selects a random sample of 50 ATM
transactions to test whether the population mean withdrawal amount has changed. The mean of the sample was $136.00. Using the
0.10 level of significance, test the claim being made using the critical value approach to hypothesis testing (5 points). Also, compute the
p-value and interpret its meaning (5 points). Explain your conclusion (5 points).
population standard deviation =
margin of error
$ 20.00
$ 3.90
α= .01
.01/2 = .005
(2.575)^2 (20^2)/ 3.9^2
6.6306 x 400 /15.21
n = 175
98.45 ± ƭ .025; 174 (15.3 /sqrt 175)
1.96 x .025
98.45 ± 2.27
Based on my conclusion at the 95% confidence
interval population mean will be
96.18 ≤ µ ≤ 100.72
The book store at IUS is conducting an end of year inventory of the text books in stock. An auditor for the store wants to estimate the mean value
of the text books in inventory at that time. She wants to have 99 percent confidence that her estimate of the average value with an error of $3.90.
Based on her experience, she estimates that the population standard deviation of the value of text books is $20.
(5 points) What sample size should be selected?
(5 points) The auditor uses the sample size determined from ‘a’, and reports the sample mean value of text books is $98.45 and the sample
standard deviation is $15.30. Using this information, construct a 95 percent confidence interval estimate for population mean.
(5 points) Based on this analysis, what is your conclusion.
Sylvania
684
831
859
893
922
939
972
1016
697
835
860
899
924
943
977
1041
720
848
868
905
926
946
984
1052
773
852
870
909
926
954
1005
1080
821
852
876
911
938
971
1014
1093
Phillips
819
907
952
994
1016
1038
1096
1153
836
912
959
1004
1018
1072
1100
1154
888
918
962
1005
1020
1077
1113
1174
897
942
986
1007
1022
1077
1113
1188
903
943
992
1015
1034
1082
1116
1230
(25 points) The length of life (in hours) of a sample of 40 100-watt light bulbs manufactured in a Sylvania plant and a sample of 40 100-watt light bulbs
manufactured in a Phillips plant are contained in tab Bulb Life in the Excel file. In order to analyze the difference in bulb life from these two manufacturers,
determine whether the variances are equal and then run the appropriate t-test to test the hypothesis that the mean bulb life of Sylvania bulbs is different than
the mean bulb life of Phillips bulbs at the .05 level of significance. State your conclusion and the basis for your conclusion.
Ho:
HA:
F.test
0.866186
8893.464
9389.874
T-Test:
P-value:
Conclusion:
2003 percentage
2003 sample size
0.80
800
2008 percentage
2008 sample size
0.68
800
As more Americans use cell phones, they question whether it is okay to talk on a cell phone while driving. Based on a survey of 800 respondents in
the year 2003 and 800 different respondents in 2008, 80% indicated it was okay to talk on a cell phone while driving in 2003 but only 68% indicated
it was okay in 2008. At the .01 level of significance, test the hypothesis that the percentage of Americans who think it’s okay to talk on a cell phone
while driving has gone down (Remember that π1 is 2003 and π2 is 2008). ). State your hypotheses (5 points), your conclusion (5 points), and the
basis for your conclusion (5 points).
H0: p1 – p2 = 0
HA: p1 – p2 ≠ 0
.8 – .68/ sqrt .74 (1-.74) (1/800 + 1 /800)
.12/ sqrt .1924 x .0025
sqrt .0005
.12/.0219
= 5.48
Based off the calculations we would reject the null hypothesis because change occurs.
Waiting Time
4.21
5.55
3.02
5.13
4.77
2.34
3.54
3.20
4.50
6.10
0.38
5.12
6.46
6.19
3.79
4.19
5.53
3.00
5.11
4.75
2.32
3.52
3.18
4.48
6.08
0.36
5.10
6.44
6.17
3.77
The average wait time at a local bank during lunch (12N to 1pm) is 5 minutes. The bank has recently implemented
procedures to reduce this wait time. Wait times were recorded over a period of two weeks and a sample of 30 wait times
in included in the tab Waittime in the Excel file. Conduct the appropriate hypothesis test using a 0.05 level of significance
to determine whether this new procedure has been successful.
(5 points) What are the null and alternate hypotheses
(5 points) What is the p-value and what does it mean?
(5 points) What is your conclusion based on this test — is the new procedure successful, or unsuccessful?
a. H0: µ ≥ 5 minutes HA: µ < 5 minutes
b. P-value -.21 ≥ -1.645 This means you do not reject the H 0.
c. Based on my conclusion the new procedure is successful
because the average time is less than 5 minutes.
Stand. Dev.
Mean
###
4.28
-0.21255753
