Question

The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100 and the standard deviation is 2. We wish to test Upper H Subscript 0 Baseline colon mu equals 100 against using a random sample of 9 specimens.

Answer 1

a)
`\alpha=P(Z<-2.25) +P(Z>2.25)=0.0122+0.0122=0.02445`

b)
`\beta = P(Z<-2.25)-P(Z<-6.75)=0.0122`

c)
`\beta= P((98.5-105)/((2)/(√(9))) < Z < (101.5-105)/((2)/(√(9))))=P(-9.75 < Z< -5.25)`

`\beta = P(Z<-5.25)-P(Z<-9.75)\approx 0`

And we can see that this value is lower compared to part b. And the reason why is because when the true mean is 105 we are far from the interval of (98.85; 101.5).

Explanation:

Part a) If the acceptance region is defined as 98.5 ≤

`\bar x` ≤ 101.5, find the type I error probability α

For this case we need to remember the definition of the significance level

`\alpha=P(Error I)=P(reject H_0 when H_o is true)`

The system of hypothesis on this case are:

Null hypothesis:
`\mu=100`

Alternatibe hypothesis:
`\mu \\eq 100`

The population deviation is given
`\sigma=2` and the sample size is n=9

The acceptance region on this case is given by:

`98.5 \leq \bar x \leq 101.5`

If we use the definition of probability of error I we got:

`\alpha= P(98.5< \bar X if \mu=100)+P(\bar X >101.5 if \mu =100)`

Using the definition of z score we got:

`\alpha=P((98.5-100)/((2)/(√(9)))< Z if \mu=100)+P((101.5-100)/((2)/(√(9)))> Z if \mu=100)`

`\alpha=P(Z<-2.25) +P(Z>2.25)=0.0122+0.0122=0.02445`

Part b: Find β for the case in which the true mean heat evolved is 103

From the definition of Type of Error II we have this:

`\beta = P(Error II) = P(Fail to reject H_o if H_o is false)`

We are assuming that the true mean is 103 so we have this:

`\beta= P((98.5-103)/((2)/(√(9))) < Z < (101.5-103)/((2)/(√(9))))=P(-6.75 < Z< -2.25)`

`\beta = P(Z<-2.25)-P(Z<-6.75)=0.0122`

Part c: c. Find β for the case where the true mean heat evolved is 105. This value of β is smaller than the one found in part (b). Why?

From the definition of Type of Error II we have this:

`\beta = P(Error II) = P(Fail to reject H_o if H_o is false)`

We are assuming that the true mean is 105 so we have this:

`\beta= P((98.5-105)/((2)/(√(9))) < Z < (101.5-105)/((2)/(√(9))))=P(-9.75 < Z< -5.25)`

`\beta = P(Z<-5.25)-P(Z<-9.75)\approx 0`

And we can see that this value is lower compared to part b. And the reason why is because when the true mean is 105 we are far from the interval of (98.85; 101.5).