8 ten thousandths as a decimal
1 answer:
You might be interested in
Answer:
D. 0.9953 (Probability of a Type II error), 0.0047 (Power of the test)
Step-by-step explanation:
Let's first remember that a Type II error is to NOT reject H0 when it is false, and the probability of that occurring is known as β. On the other hand, power refers to the probability of rejecting H0 when it is false, so it can be calculated as 1 - β.
To resolve this we are going to use the Z-statistic:
Z = (X¯ - μ0) / (σ/√n)
where μ0 = 1.2
σ = 5
n = 42
As we can see in part A of the attached image, we have the normal distribution curve representation for this test, and because this is a two-tailed test, we split the significance level of α=0.01 evenly into the two tails, 0.005 in each tail, and if we look for the Z critical value for those values in a standard distribution Z table we will find that that value is 2.576.
Now we need to stablish the equation that will telll us for what values of X¯ will we reject H0.
Reject if:
Z ≤ -2.576 Z ≥ 2.576
We know the equation for the Z-statistic, so we can substitute like follows and resolve.
Reject if:
(X¯ - 1.2) / (5/√42) ≤ -2.576 (X¯ - 1.2) / (5/√42) ≥ 2.576
X¯ ≤ -0.79 X¯ ≥ 3.19
We have the information that the true population mean is 1.25, so we now for a fact that H0 is false, so with this we can calculate the probability of a Type II error: P(Do not reject H0 | μ=1.25)
As we can see in part B of the attached image, we can stablish that the type II error will represent the probability of the sample mean (X¯) falling between -0.79 and 3.19 when μ=1.25, and that represents the shaded area. So now we now that we are looking for P(-0.79 < X¯ < 3.19 | μ=1.25).
Because we know the equation of Z, we are going to standardize this as follows:
P ( (-0.79 - 1.25) / (5/√42) < Z < (3.19 - 1.25) / (5/√42) )
This equals to:
P(-2.64 < Z < 2.51)
If we go and look for the area under the curve for Z positive scores in a normal standart table (part C of attached image), we will find that that area is 0.9940, which represents the probability of a Type II error.
Therefore, the power of the test will be 1-0.9940 = 0.006
If we look at the options of answers we have, there is no option that looks like this results, which means there was a probable redaction error, so we are going to stay with the closest option to these values which is option D.
The probability that the SRS of 10 students will spend an average of between 600 and 700 dollars is 0.8132.
Let x be the total amount spent by students.
x follows normal distribution with mean μ = 650,
standard deviation δ = 120
We take a simple random sample of size n = 10
We are asked to find average spending of 10 students ( ) is between 600 and 700
We have to find P( 600 <= <= 700)
According to the sampling distribution of the sample mean , it follows an approximately normal distribution with mean μ{
} = μ and standard deviation δ{
} = {δ}/{√n}
Therefore here mean of ,( μ{
} ) = 650 and
standard deviation of , δ{
} = {120}/{√10} = 37.9473
The probability that the SRS of 10 students will spend an average of between 600 and 700 dollars is,
Let Z= x - μ / δ
Z₁ = 600 - 650 / 37.9473 = -1.32 similarly
Z₂ = 700 - 650 / 37.9473 = 1.32
From standard normal distribution table, P( -1.32 < < 1.32) = 0.8132
The probability that the SRS of 10 students will spend an average of between 600 and 700 dollars is 0.8132
To learn more about Normal distribution click here:
#SPJ4