Reference no: EM13926838
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
Degrees of freedom = n- 1 = 36 -1 = 35
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
Since it says that that we to find evidence that cell phone use decreased from 1996 to 1997, there is one tailed test and the null hypothesis is that the mean is equal to 42.40
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
If Null hypothesis is as from above, the alternative is mean greater than 42.40
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
From the above, the sample mean outcome is 45.70
9 What is the DSM Standard deviation?
A 5
B 42.40
C 8.95
D 22
E 3.44
DSM standard deviation = Standard deviation/ sqrt(n) = 22/sqrt(41) = 3.44
10 What are the degrees of freedom?
A 40
B 21
C 4
D 8
E There are no degrees of freedom needed for this example
Since n > 30, we will use Z-test and hence there are no degrees of freedom
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
Since it's a one-tailed test, we will have alpha = 5%
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
Using Z -test = (42.40 - 45.70)/3.44 = - 0.96
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.
At Z = -0.96, p-value = 0.1685. since p-value is greater than 5% level of significance, we reject the null hypothesis
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 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
DSM standard deviation = Standard deviation/ sqrt(n) = 5/sqrt(16) = 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