Answer To: Need clear and simple solutions to help me to study for stats test. Document Preview: ECMT1010 –...
Robert answered on Dec 21 2021
A1) We will construct the test statistic first using the given information. The test statistic is given by:
n
x
Z
/
Using the given values , we compute Z under H0 . Thus, Z=-2.3078
Now the critical value at 1% and one sided is 2.32635 (Using standard normal tables)
The modulus Z is less than the critical value, therefore we do not reject the null hypothesis
that 48.7 .
A2) Null hypothesis,H0: 82600
Alternative hypothesis,H1: 82600
Where is the average income of BMW driver
Test statistic:
n
x
Z
/
7810,18,78974 nx
Thus Under H0,Z=-1.969
The critical value at 1% and one sided is 2.32635
Since the modulus value of Z is less than 2.32635, we do not reject the null hypothesis at 1%
level of significance and conclude that there is insufficient evidence to reject that average income of
BMW driver is equal to 82600.
A3)In this question, since we do not know the population standard deviation we would use the t test
instead of Z.
The test statistic is given by:
ns
x
t
/
, where s is the sample standard deviation.
From the values given in the question, we have 59.3,20,45.16 snx
Thus t=0.5605
The critical value for t with n-1 degrees of freedom is :2.093.
Again since the observed is less than tabulated value , we do not reject the null hypothesis at 5% ,
two tailed test.
A4)Setting up the null and alternative hypothesis first.
H0: 80.322
H1: 80.322
Where is the average price per sq metre.
Test statistic:
ns
x
t
/
From the information given , 90.12,19,70.316 snx
Thus , t=-2.061184
From the t tables, critical value is 2.1, (18 d.f)
Since the calculated value of modulus t is less than 2.1, we do not reject the null hypothesis
at 5 % level .
A5)We first compute the test statistic for this proportion problem.
The test statistic is
n
pp
pp
Z
)1(
ˆ
Where 28.0740/207ˆ p is the sample proportion and n=740
Thus Under H0, Z=-0.599
The critical values at 5% two sided are: -1.96 and 1.96.
If Z lies within the range then the null hypothesis is not...