Question

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IS 25 1) CL E) 10) (이 13)

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Answer 1

Note : Final answers are highlighted in colour.

1) Probability model is Normal population.

2)

Null hypothesis is Delivery time is atleast 37 hours

Alternate hypothesis is Delivery time is leass than 37 hours

Therefore,

Hypothesis:

H₀:

011 AI 37

H_a:

HO <37

Left alternative test

3)As we have the population standard deviation($\sigma$), we need to use Z stat(Normal distribution)

Therefore test statistic(Z stat)= $\frac{\overline{X}-\mu}{\sigma/\sqrt{n}}$ ~N(0,1)

Two

4)Sampling distribution is exact as the delivery times are normally distributed and we have the population standard deviation as well. We dont need to resort to central limit theorem.

Exact

5)

Z= $\frac{\overline{X}-\mu}{\sigma/\sqrt{n}}$ = $\frac{35-37}{8/\sqrt{64}}$ = -2

.00 .01 102 03 .04 405 . 06 407 . 08 .09 3.4 3.3 -3.2 -3.1 -3.0 -29 -2.8 -2.7 -2.6 -2.5 -2.4 -2.3 -2.2 -2.1 -2.0 10003 0005 10007 10010 ,0013 10019 10026 10035 0047 10062 20082 20107 .019 .019 0228 10003 20003 10005 .0005 ,0007 1.0006 2009 .0009 20013 20013 2018 10018 10025 .0024 004 .0033 10045 ,004 .2050 .0059 20080 20078 2014 20102 .016 .0132 .0174 .0170 00222 1.0217 10003 .0004 .0006 .0009 1.0012 .0017 .0023 .0032 ,0043 20057 10075 .009 .0129 .0166 0212 10003 .0004 .0006 10008 1.0012 .0016 .0023 .0031 10041 20055 .0073 20096 .0125 .0162 0207 0003 .0004 .0006 .0008 .0011 20016 0022 .0030 0040 20054 20071 .0094 20122 .0158 .0202 10003 .0004 .0006 40008 .0011 .0015 .0021 20029 1.0039 .0052 .0069 20091 .0119 .0154 .0197 10003 10004 ,0005 10008 1.0011 20015 20021 10028 40038 20051 20068 .009 10116 .0150 10192 10003 .0004 ,0005 10007 20010 10014 .0020 10027 .0037 004中 20066 100B) 0113 1016 0198 1,0002 10003 ,0005 10007 10010 10014 1.0019 .0026 .0036 10048 20064 10024 .0110 20143 20183

P(Z<-2)= 0.0228(From the Z table above)

6)$\mu$=37 means the null hypothesis is true, we are trying to find out the probability of rejecting the null hypothesis when it is true. This is called Type 1 error.

Probability of Type 1 error is called level of significance ($\alpha$).

7)Critical value= 35

P(H₀ will be accepted if $\mu$ =34)= P($\overline{X}$>35 | $\mu$ =34)

Z = $\frac{\overline{X}-\mu}{\sigma/\sqrt{n}}$ = $\frac{35-34}{8/\sqrt{64}}$ =1

P($\overline{X}$>35 | $\mu$ =34)= P(Z>1)=  0.158655

8) The calculated is P(Not rejecting H₀ | H₀ is false).

We are trying to calculate the probability of not rejecting H₀ when it is to be rejected. This is called Type II error($\beta$).

11) Sample mean($\overline{X}$)= 35.5

Z= $\frac{35.5-37}{8/\sqrt{64}}$ = -1.5

P(Z<-1.5)= 0.0668

12)critical vale when $\alpha$ =0.01,

Z value for  $\alpha$=0.01 is -2.326

corresponding X= $\mu+Z*\frac{\sigma}{\sqrt{n}}$ = 37-2.326*1= 34.67

13)Power = 1-$\beta$= 0.95

$\beta$= 0.05,  $\alpha$=0.01

Corresponding Z for $\alpha$ =0.01 is -2.326

X= 37- 2.326*8/

n

n
= $\frac{-2.326*8}{X-37}$ = -18.608/(X-37)---------------------(1)

P(H₀ will be accepted if $\mu$ =34)= 0.05

$\frac{X-\mu}{\sigma/\sqrt{n}}$= 0.05

X-34= 0.05*$\frac{8}{\frac{-18.608}{X-37}}$

Solving for X from the above equation gives, X= 34.06.

Solving for n from (1) gives n= 40

The minimum sample size =40