1.A.1. The null hypothesis is
0 : 6H (The population mean is 6 days)
Against the alternative hypothesis
1 : 6H (The population mean is less than 6 days)
Use 0.05 in one tail.
The mean and standard deviation of 25 observations are 5.648 days (M) and 1.194 days (S).
The degrees of freedom is-
1
25 1
24
n
Calculate the estimated standard error-
2
21.194
25
0.2388
M
s
S
n
Now, calculate the t- statistic
5.648 6
0.2388
1.47
M
M
t
S
From t- table, we find the critical value of 0.05,24 1.711t ,
The test statistic does not fall in the critical region, so fail to reject the null hypothesis at 0.05
level. Thus, it can be concluded that the test is not significant.
A2. Go to DDXL, select 1 Var t test, enter 6 in the hypothesized mean box, select alpha=0.05,
select μ<μ0, click compute. The final output is as follows:
A.3. The level of significance is 0.05
The null and alternative hypotheses:
0 1: 2, : 2H H
The value of chi square statistic is:
2
2
2
0
2
2
( 1)
(15 1) * 2.712
2
25.74
N s
The degrees of freedom=N-1=15-1=14
About the original distribution we assume a normal population distribution.
The critical value is chi square(14,0.05)=23.685, the test statistic falls in the critical region so
reject the null hypothesis.
Go to DDXL, click hypothesized tests, check chi square for sd. Enter the hypothesized value,
select alpha, select the direction of the hypothesis. The final output is as follows.
B1. The null hypothesis is
0 : 0.95H P (The percentage of population proportion is 95%)
Against the alternative hypothesis
1 : 0.95H P (The percentage of population proportion is less than 95%)
Thus, calculate the sample proportion
562
=
600
=0.937
x
p
n
Calculate the standard deviation
1
=
0.937 0.063
=
600
=0.01
p p
n
Now, calculate the z statistic:
0.937 0.95
0.01
=-1.3
p P
z
Now, with 0.05 , the critical value of z is minus 1.96.
Since, the test statistic does not fall in the critical region so fail to reject the null hypothesis.
Thus, the test is not significant at 0.05 level of significance.
Go to DDXL, select hypothesis testing, select 1 var prop test. Enter A1 in number of success (562)
and enter B1 in number of trials (600). Set p0=0.95, set alpha=0.05, select alternative p
final output is as follows:
II. The data set is given below:
Using Excel find the following values:
1 1 2 121.1, 1.421, 22.75, 3.202x s x s
Calculate the pooled variance:
2 21 1 2 22
1 2
2 2
1 ( 1)
2
(5 1) *1.421 (4 1) *3.202
=
5 4 2
=5.55
p
n s n S
S
n n
Now, calculate the estimated standard error
1 2
2 2
1 2
5.55 5.55
=
5 4
=1.58
p...