What is 0.000303 in scientific notation
1 answer:
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(a) Animals by popularity; Dogs, Cats, Others, Horses.
(b) If 250 students were interviewed, 20 students have other pets.
(c) No, it was incorrect.
Step-by-step explanation:
Step 1; We arrange the different pets in chronological order, from the highest percent to the lowest percent;
1. Dogs - 67%,
2. Cats - 20%
3. Others - 8%,
4. Horses - 1%.
None is not included in this list.
Step 2; If 250 students were interviewed, 8% of those have other pets. So we need to calculate how much 8% of 250 is.
8% of 250 = × 250 = 0.08 × 250 = 20.
So 20 students had other pets.
Step 3; 20% of the interviewed students had cats. If 75 students had cats, we need to equate 20% to the 75 and find out how many students were actually interviewed.
20% × x = 75, 0.2 × x = 75,
x = = 375.
So if 375 students were interviewed, 75 students would have cats. So the estimate of 250 students being interviewed is incorrect.
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.
