Thread: Probability of a Type II error

1. Probability of a Type II error

I can't seem to figure out how to obtain the probability of type 2 errors from this question.

A researcher is conducting a hypothesis test to determine if the true mean weight of all babies born at a particular hospital exceeds 7.8 pounds. He takes a sample of 15 newborn babies and weighs them. Weights are known to follow a normal distribution with a standard deviation of 1.4 pounds.

What is the probability of a Type II error if the true mean weight is actually 8.1 pounds?

Any help greatly appreciated!

2. Originally Posted by bobby big dog malone
I can't seem to figure out how to obtain the probability of type 2 errors from this question.

A researcher is conducting a hypothesis test to determine if the true mean weight of all babies born at a particular hospital exceeds 7.8 pounds. He takes a sample of 15 newborn babies and weighs them. Weights are known to follow a normal distribution with a standard deviation of 1.4 pounds.

What is the probability of a Type II error if the true mean weight is actually 8.1 pounds?

Any help greatly appreciated!
Assuming that your prespecified Type I error rate is 0.05 I get:

Pr{Type II error}=0.792.

3. How?

Originally Posted by Dragan
Assuming that your prespecified Type I error rate is 0.05 I get:

Pr{Type II error}=0.792.
Can you please elaborate a little on how you get this?

4. Originally Posted by SVM

Can you please elaborate a little on how you get this?
Sure, under the null hypothesis we have:

H[0]: Mu=7.8.

Because we are only considering the right hand side of the normal distribution this gives a critical value (alpha 0.05) associated with the mean of:

XBar = zcrit*StdError + Mu
XBar = 1.6445*(1.4/Sqrt[15]) + 7.8 =8.394632....

However, the true mean is Mu=8.1.

Thus,
zcrit = (8.394632 - 8.1) / (1.4/Sqrt[15]) = 0.815075...

Therefore, the probability of making a Type II error will be:

Pr{Type II eror} = Pr{Z<=z=0.815075}=0.792485...

from using the standard normal cumulative distribution function (cdf).

Note that the Power would be: 1 - Type II error = 0.1849...

5. how do you find z = 1.6445 and Pr{Type II error} = Pr{Z<=z=0.815075}=0.792485?

6. Originally Posted by bobby big dog malone
how do you find z = 1.6445 and Pr{Type II error} = Pr{Z<=z=0.815075}=0.792485?
For both cases you would use the standard normal cumulative distribution function (cdf). You can look these values up in statistics textbook if you don't have software to compute them (Minitab) - however Excel does do this. I used Minitab to obtain both values.

Now, in terms of z=1.6445 I assumed that your Type I error rate was 0.05 - which is what is usually done.

Thus,

Pr{Z>=z=1.6445} = 1 - Pr{Z<=z=1.6445} = 0.05

7. Thanks allot. I just have a new question that I need some guidance with.

The label on a can of a particular brand of canned peaches says the can contains 16 ounces of fruit. According to the company, the weight of peaches per can follows a normal distribution with a standard deviation of 0.4 ounces. You would like to conduct a hypothesis test at α = 0.025 to determine if the mean weight is lower than the advertised weight.

a) You buy 7 cans of peaches and weigh the contents of each can. What is the power of the test if the true mean weight of the peaches is actually 15.75 ounces per can? Answer: 0.380

I know you have to first find the probability of a type 2 error then subtract that from 1 to find the power of the test I just don't know how their coming to an answer of 0.380.

8. Originally Posted by bobby big dog malone
Thanks allot. I just have a new question that I need some guidance with.
.......
I just don't know how their coming to an answer of 0.380.

You would use the same approach as I described above. It's just that you have to be careful because your on the left-hand side of the normal distribution.

Specifically,

Once you obtain z = -0.306405 you would then find power as

Pr{Z<=z=-0.306405} = 0.379648~0.380.

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