# Project 6 Math 102 | Complete Solution

Project 6 Math 102
1.When a directional hypothesis is being tested and the rejection region is on the right of the mean, then the alternate hypothesis must be of the form:
A Less than
B Greater than
C Not equal
D Equal
2 As the significance level decreases the chances of rejecting
increases.
A True
B False
C Cannot determine
D Stays the same
3 A directional alternate hypothesis results in a two tail test.
A True
B False
C Cannot determine
D Depends on the alpha
4 If samples of size 36 are selected in a one sample hypothesis test using t-scores, then what would be the degrees of freedom?
A 36
B 37
C 35
D 34
5-13
In 1996 the mean monthly bill for cell phone users was \$45.70 with a standard deviation of \$22. In 1997 the mean monthly bill for cell phone use for a sample of 41 phone bills was \$42.40. At the 5% significance level, is there sufficient evidence to conclude that the cell phone use
decreased from 1996 to 1997? )( Assume the population is a Normal Population.)
5 What would be the Null Hypothesis?
A μ = 42.40
B μ < 45.70
C μ = 45.70
D μ 42.40
E μ 45.70
6 What is the Alternate Hypothesis?
A μ = 42.40
B μ < 45.70
C μ = 45.70
D μ 42.40
E μ 45.70
7 What is the given standard deviation?
A 5
B 42.40
C 45.70
D 22
E 3.44
8 What is the sample mean outcome?
A 5
B 42.40
C 45.70
D 22
E 3.44
9 What is the DSM Standard deviation?
A 5
B 42.40
C 8.95
D 22
E 3.44
10 What is the degrees of freedom?
A 40
B 21
C 4
D 8
E There are no degrees of freedom needed for this example
11 What is the value of the critical cut off score?
A - 1.65
B -2.33
C -1.68
D 1.68
E 1.65
12 What is the values of the test statistic?
A -0.96
B -1.65
C -1.68
D 0.96
E None of the above
13 What is the final decision?
A Do Not Reject Ho
B Do Not Reject Ha
C Reject Ho
D Reject Ha
E Neither Reject or Do not Reject, since it is a tie.
14-
22 An automobile manufacturer claims that their leading compact car averages 32 mpg or
more in the city. A local police department purchased 16 of these cars. The cars were
driven exclusively in the city and averaged 28.2 mpg with a standard deviation of 5 mpg.
Is the manufacturer’s claim too high? (Use a = 1% )( Assume the population is a
Normal Population.)
14 What would be the Null Hypothesis?
A μ = 32
B μ < 32
C μ = 28.2
D μ 28.2
E μ = 5
15 What is the Alternate Hypothesis?
A μ = 32
B μ < 32
C μ < 28.2
D μ 28.2
E μ 32
16 What is the given standard deviation?
A 2.5
B 5
C 16
D 14.1
E 8
17 What is the sample mean outcome?
A 35
B 16
C 5
D 28.2
E 32
18 What is the DSM Standard deviation?
A 2.5
B 16
C 5
D 8
E 1.25
19 What is the degrees of freedom?
A 16
B 31
C 4
D 15
E There are no degrees of freedom needed for this example
20 What is the value of the critical cut off score?
A - 1.75
B -2.60
C +1.75
D -2.13 and 2.13
E 2.60
21 What is the values of the test statistic?
A -3.04
B -1.4
C 5
D 28.2
E 1.4
22 What is the final decision?
A Do Not Reject Ho
B Do Not Reject Ha
C Reject Ho
D Reject Ha
E Neither Reject nor Do not Reject, since it is a tie.
23-
31 According to the US Bureau of Labor Statistics, the average consumer spent \$1729 on
apparel and services in 1997. That same year, 31 consumers in the Midwest spent a
mean amount of \$1552 with a standard deviation of \$362. At the 5% level, do the data
provide sufficient evidence to conclude that the 1997 expenditures for consumers in the
Midwest were significantly different from the national mean of \$1729? )( Assume the
population is a Normal Population.)
23 What would be the Null Hypothesis?
A μ = 1552
B μ not equal 1552
C μ = 28.2
D μ not equal 1729
E μ = 1729
24 What is the Alternate Hypothesis?
A μ = 1552
B μ not equal 1552
C μ = 28.2
D μ not equal 1729
E μ = 1729
25 What is the given standard deviation?
A 362
B 65.02
C 31
D 1.63
E 0.05
26 What is the sample mean outcome?
A 1729
B 362
C 1552
D 31
E 5
27 What is the DSM Standard deviation?
A 65.02
B 362
C 1.63
D 278.75
E 9626.66
28 What is the degrees of freedom?
A 361
B 30
C 4
D 31
E There are no degrees of freedom needed for this example
29 What is the value of the critical cut off score?
A - 1.70 and 1.70
B 2.04
C -2.04 and 2.04
D -1.70
E -2.75 and 2.75
30 What is the value of the test statistic?
A -3.5
B 2.72
C 3.5
D -2.04
E -2.72
31 What is the final decision?
A Do Not Reject Ho
B Do Not Reject Ha
C Reject Ho
D Reject Ha
E Neither Reject or Do not Reject, since it is a tie.
32-
40
The Food and Nutrition Board of the National Academy of Sciences recommends a daily
allowance of calcium for adults of 790 mg. A random sample of 38 people with incomes
below the poverty level had a mean daily intake of 747.7 mg of calcium. At the 1%
significance level, do the data provide evidence to conclude that the calcium intake of
people with incomes below the poverty level is less than the recommended allowance of
790 mg? Assume that = 210 mg. )( Assume the population is a Normal Population.)
32 What would be the Null Hypothesis?
A μ = 747.7
B μ < 747.7
C μ = 210
D μ 790
E μ = 790
33 What is the Alternate Hypothesis?
A μ = 747.7
B μ < 790
C μ = 210
D μ 790
E μ = 790
34 What is the given standard deviation?
A 1
B 210
C 790
D 34.07
E 2.62
35 What is the sample mean outcome?
A 210
B 790
C 747.7
D 38
E 121.29
36 What is the DSM Standard deviation?
A 1
B 210
C 790
D 34.07
E 2.62
37 What is the degrees of freedom?
A 37
B 209
C 38
D 39
E There are no degrees of freedom needed for this example
38 What is the value of the critical cut off score?
A - 1.68
B -2.33
C -2.42
D -2.58
E 1.68
39 What is the values of the test statistic?
A -3.8
B -16.13
C -.01
D 16.13
E -1.24
40 What is the final decision?
A Do Not Reject Ho
B Do Not Reject Ha
C Reject Ho
D Reject Ha
E Neither Reject or Do noeject, since it is a tie.
41-
49The average retail price for bananas in 1998 was 49.0 cents per pound, as reported by
the US Department of Agriculture. Recently, a random sample of 25 markets had the
mean price for bananas of 52.4 cents per pound with a standard deviation of 3.8 cents.
Can you conclude that the current mean retail price for bananas is significantly higher at
the 5% level of significance? )( Assume the population is a Normal Population.)
41 What would be the Null Hypothesis?
A μ = 52.4
B μ < 52.4
C μ =25
D μ 49
E μ = 49
42 What is the Alternate Hypothesis?
A μ = 52.4
B μ < 52.4
C μ =25
D μ 49
E μ = 49
43 What is the given standard deviation?
A 0.76
B 3.8
C 9.8
D 10.48
E 25
44 What is the sample mean outcome?
A 52.4
B 25
C 49
D 3.8
E 5
45 What is the DSM Standard deviation?
A 10.48
B 9.8
C 3.8
D 0.76
E 25
46 What is the degrees of freedom?
A 24
B 48
C 4
D 51
E There are no degrees of freedom needed for this example
47 What is the value of the critical cut off score?
A - 1.71
B 1.71
C 1.65
D 2.06
E 2.49
48 What is the values of the test statistic?
A -4.47
B 2.58
C -2.58
D 4.47
E 0.89
49 What is the final decision?
A Do Not Reject Ho
B Do Not Reject Ha
C Reject Ho
D Reject Ha
E Neither Reject or Do not Reject, since it is a tie.
50
Having a baby in a US hospital costs \$2628, on average, with a standard deviation of \$456. Active management of labor (AML) is a group of interventions designed to help reduce the length of labor and the rate of cesarean deliveries. Physicians were interested in determining whether AML would reduce the cost for delivery. So they researched the cost of 200 AML deliveries and found a mean cost of \$2498. At the 1% significance level, do the data provide sufficient evidence to conclude that, on average, AML reduces the cost of having a baby in the US hospitals? )( Assume the population is a Normal Population.)
50
The above hypothesis test would make use of which statistic?
a) Normal z scores b) T – scores c) C – scores d) r scores
End of Project 7