In this Lab you will be 8 completed test review questions. It is your turn to be the evaluator. Check each problem to see if they are correct or not. If there is a mistake(s) please make corrections....

1)
Given that
S =$17.54
n = 160
We want to test the hypothesis that the mean cost of a spreading ticket is significantly increased than $150.00 in year 2002
Null hypothesis(H0): the mean cost of a spreading ticket is not significantly increased than $150.00 in year 2002
Alternative hypothesis(H1): the mean cost of a spreading ticket is significantly increased than $150.00 in year 2002
Level of Significance value(α) =0.01
Test statistic: Under H0, The test statistic formulae can be defined
=2.88
Using Calculator we find Critical value
Tcritical
= INVT(0.01,159)
=2.35
Using Calculator we find P-value
= 0.0023
Decision: Here we observe that P-value(0.0023) is less than level of significance value(0.01).So we reject the null hypothesis(H0)
Conclusion: Therefore we conclude that there is a sufficient evidence to support that the police department’s claim that the mean cost of a speeding ticket is significantly increased than $150.00 in year 2002
2)
Given that
14.1
S =1.9
n = 20
We want to test the hypothesis that the new mean is significantly different from 15 minutes
Null hypothesis(H0): The new mean is not significantly different from 15 minutes
Alternative hypothesis(H1): The new mean is significantly different from 15 minutes
Level of Significance value(α) =0.05
Test statistic: Under H0, The test statistic formulae can be defined
=-2.12
Degrees of Freedom (df) = n-1
= 20-1
=19
Using Calculator we find Critical value
Tcritical
= INVT(0.05,19)
=2.09
Using Calculator we find P-value
= 0.0474
Decision: Here we observe that P-value(0.000474) is less than level of significance value(0.05).So we reject the null hypothesis(H0)
Conclusion: Therefore we conclude that there is a sufficient evidence to support that the mean installation time is significantly different from 15 minutes.
3)
Given that
n =250
x =20
We want to test the whether the Cost effective is more than 15% of the alumini and supporters provide monetary contributions.
Null And Alternative hypothesis are
H0: p
H1: p
Level of significance value
Critical value using calculator
Zcritical = -1.645
P-value using Calculator
P-value = 0.3300
Decision: Here we observe that P-value(0.3300) is greater than level of significance value(0.05).So we fail to reject the null...