QUESTION 18 A carpet company advertises that it will deliver your carpet within

Question

QUESTION 18 A carpet company advertises that it will deliver your carpet within 15 days of purchase. A sample of 49 past customers is taken. The average delivery time in the sample was 16.6 days. Assume the population standard deviation is known to be 5.6 days. Answer the following questions: . What is the cut-offpoint at alpha = .05? 2. What is the cut-offpoint ifType I error is 1%? 3, Would you reject the Null at 5% level of significance? 4. Would you reject the Null at 1% level of significance? 5. If suppose the true delivery time for the population is 17 days. What is the Type II error at Alpha .01? 2Ss

Explanation / Answer

18.
1).at level of significance = 0.05
Given that,
population mean(u)=15
standard deviation, =5.6
sample mean, x =16.6
number (n)=49
null, Ho: =15
alternate, H1: !=15
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 16.6-15/(5.6/sqrt(49)
zo = 2
| zo | = 2
critical value
the value of |z | at los 5% is 1.96
we got |zo| =2 & | z | = 1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 2 ) = 0.0455
hence value of p0.05 > 0.0455, here we reject Ho
ANSWERS
---------------
null, Ho: =15
alternate, H1: !=15
test statistic: 2
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.0455

2)
At level of significance = 0.01 Type 1 error
Given that,
Standard deviation, =5.6
Sample Mean, X =16.6
Null, H0: =15
Alternate, H1: !=15
Level of significance, = 0.01
From Standard normal table, Z /2 =2.58
Since our test is two-tailed
Reject Ho, if Zo < -2.58 OR if Zo > 2.58
Reject Ho if (x-15)/5.6/(n) < -2.58 OR if (x-15)/5.6/(n) > 2.58
Reject Ho if x < 15-14.448/(n) OR if x > 15-14.448/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 49 then the critical region
becomes,
Reject Ho if x < 15-14.448/(49) OR if x > 15+14.448/(49)
Reject Ho if x < 12.936 OR if x > 17.064
Suppose the true mean is 17
Probability of Type I error,
P(Type I error) = P(Reject Ho | Ho is true )
= P(12.936 < x OR x >17.064 | 1 = 17)
= P(12.936-17/5.6/(49) < x - / /n OR x - / /n >17.064-17/5.6/(49)
= P(-5.08 < Z OR Z >0.08 )
= P( Z <-5.08) + P( Z > 0.08)
= 0 + 0.4681 [ Using Z Table ]
= 0.468

3)
At level of significance = 0.05
Yes, Reject the null hypothesis,done in (1) problem

4)
At level of significance = 0.01
Given that,
population mean(u)=15
standard deviation, =5.6
sample mean, x =16.6
number (n)=49
null, Ho: =15
alternate, H1: !=15
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.58
since our test is two-tailed
reject Ho, if zo < -2.58 OR if zo > 2.58
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 16.6-15/(5.6/sqrt(49)
zo = 2
| zo | = 2
critical value
the value of |z | at los 1% is 2.58
we got |zo| =2 & | z | = 2.58
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 2 ) = 0.0455
hence value of p0.01 < 0.0455, here we do not reject Ho
ANSWERS
---------------
null, Ho: =15
alternate, H1: !=15
test statistic: 2
critical value: -2.58 , 2.58
decision: do not reject Ho
p-value: 0.0455
No, reject Ho null hypothesis

5)
Given that,
Standard deviation, =5.6
Sample Mean, X =16.6
Null, H0: =15
Alternate, H1: !=15
Level of significance, = 0.01
From Standard normal table, Z /2 =2.5758
Since our test is two-tailed
Reject Ho, if Zo < -2.5758 OR if Zo > 2.5758
Reject Ho if (x-15)/5.6/(n) < -2.5758 OR if (x-15)/5.6/(n) > 2.5758
Reject Ho if x < 15-14.42448/(n) OR if x > 15-14.42448/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 49 then the critical region
becomes,
Reject Ho if x < 15-14.42448/(49) OR if x > 15+14.42448/(49)
Reject Ho if x < 12.93936 OR if x > 17.06064
Implies, don't reject Ho if 12.93936 x 17.06064
Suppose the true mean is 17
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(12.93936 x 17.06064 | 1 = 17)
= P(12.93936-17/5.6/(49) x - / /n 17.06064-17/5.6/(49)
= P(-5.0758 Z 0.0758 )
= P( Z 0.0758) - P( Z -5.0758)
= 0.5302 - 0 [ Using Z Table ]
= 0.5302
For n =49 the probability of Type II error is 0.5302

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