Respuesta :
Answer:
There is a probability of P=0.02 of making a Type II error if the true mean is μ=1130.
Step-by-step explanation:
This is an hypothesis test for the lifetime of a certain ype of light bulb.
The population distribution is normal, with mean of 1,000 hours and STD of 110 hours.
The sample size for this test is n=10.
The significance level is assumed to be 0.05.
In this case, when the claim is that the new light bulb model has a longer average lifetime, so this is a right-tailed test.
For a significance level, the critical value (zc) that is bound of the rejection region is:
[tex]P(z>z_c)=0.05[/tex]
This value of zc is zc=1.645.
This value, for a sample with size n=10 is:
[tex]z_c=\dfrac{X_c-\mu}{\sigma/\sqrt{n}}\\\\\\X_c=\mu+\dfrac{z_c\cdort\sigma}{\sqrt{n}}=1000+\dfrac{1.645*110}{\sqrt{10}}=1000+57.22=1057.22[/tex]
That means that if the sample mean (of a sample of size n=10) is bigger than 1057.22, the null hypothesis will be rejected.
The Type II error happens when a false null hypothesis failed to be rejected.
We now know that the true mean of the lifetime is 1130, the probability of not rejecting the null hypothesis (H0: μ=1100) is the probability of getting a sample mean smaller than 1057.22.
The probability of getting a sample smaller than 1057.22 when the true mean is 1130 is:
[tex]z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{1057.22-1130}{110/\sqrt{10}}=\dfrac{-72.78}{34.7851}=-2.0923 \\\\\\P(M<1057.22)=P(z<-2.0923)=0.01821[/tex]
Then, there is a probability of P=0.02 of making a Type II error if the true mean is μ=1130.
Using the normal distribution and the central limit theorem, it is found that there is a 0.0001 = 0.01% probability of a type II error.
In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
In this problem:
- Mean of 1130 hours, hence [tex]\mu = 1130[/tex]
- Standard deviation of 110 hours, hence [tex]\sigma = 110[/tex]
- Sample of 10 bulbs, hence [tex]n = 10, s = \frac{110}{\sqrt{10}}[/tex].
We test if the average lifetime is longer, and a Type II error is concluding that it is not longer when in fact it is longer, hence, it is the probability of finding a sample mean below 1000 hours, which is the p-value of Z when X = 1000.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{1000 - 1130}{\frac{110}{\sqrt{10}}}[/tex]
[tex]Z = -3.74[/tex]
[tex]Z = -3.74[/tex] has a p-value of 0.0001.
0.0001 = 0.01% probability of a type II error.
A similar problem is given at https://brainly.com/question/15186499
