Answer:
d. Decrease
Step-by-step explanation:
A Type II error is when we fail to reject a false null hypothesis. Higher values of α make it easier to reject the null hypothesis, so choosing higher values for α can reduce the probability of a Type II error.
The consequence here is that if the null hypothesis is true, increasing α makes it more likely that we commit a Type I error (rejecting a true null hypothesis).
So using lower values of α can increase the probability of a Type II error.
Raising the level of significance in a hypothesis test from .01 to .05 would decrease the probability of making a Type II error. This is because as we become more accepting of risk in making a Type I error, we simultaneously reduce the risk of making a Type II error.
The level of significance in a hypothesis test is the probability that we are willing to accept for incorrectly rejecting the null hypothesis or making a Type I error. If the level of significance is raised, there is a higher chance we incorrectly reject the null hypothesis, increasing the chances of a Type I error. However, this also has an effect on the probability of committing a Type II error, which is to incorrectly accept the null hypothesis.
Specifically, when the level of significance of a hypothesis test is raised from .01 to .05, the probability of a Type II error (option b) will decrease. The reason for this is that increasing the level of significance or alpha means we are more likely to reject the null hypothesis. As we are more accepting of risk in terms of making a Type I error, we are less likely to make a Type II error, as the two error types often move in opposite directions. Thus, the answer to your question is d. The probability of a Type II error will decrease if the significance level is raised from .01 to .05.
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Answer:
The value is -55
In the equation x-10=-15, x is equal to -5.
Therefore, the value of x is plugged into the second equation and the value found is -55.
Answer:
x = -55
Step-by-step explanation:
We know that x - 10 = -15 and we want to find the value of 5x - 30 but our first step would be to find the value x so then we can can substitute that back into the expression so
x - 10 = -15
⇔ Add -10 to both sides to isolate x
x = -5
So now we know that value of 'x' is -5 we can substitute it back into the expression 5x - 30 so
5x - 30
→ Substitute x = -5 back into it
5 × -5 - 30
→ Simplify
-25 - 30 = -55
If x - 10 = -15, then find the value of 5x - 30 the value of x is -55
b. P(28c. P(X>35)
d. P(X>31)
e. the mileage rating that the upper 5% of cars achieve.
The upper 5% of cars have a mileage rating of 35.805 mpg
Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:
z = (raw score - mean) / standard deviation
Given; mean of 33 mpg and a standard deviation of 1.7
a) For < 30:
z = (30 - 33)/1.7 = -1.76
P(x < 30) = P(z < -1.76) = 1 - 0.8413 = 0.0392
b) For < 28:
z = (28 - 33)/1.7 = -2.94
P(x < 28) = P(z < -2.94) = 0.0016
c) For > 35:
z = (35 - 33)/1.7 = 1.18
P(x > 35) = P(z > 1.18) = 1 - P(z < 1.18) = 1 - 0.8810 = 0.119
d) For > 31:
z = (31 - 33)/1.7 = -1.18
P(x > 31) = P(z > -1.18) = 1 - P(z < -1.18) = 0.8810
e) The upper 5% of cars achieve have a z score of 1.65, hence:
1.65 = (x - 33)/1.7
x = 35.805 mpg
The upper 5% of cars have a mileage rating of 35.805 mpg
Find out more on z score at: brainly.com/question/25638875
Answer:
a) P(X < 30) = 0.0392.
b) P(28 < X < 32) = 0.2760
c) P(X > 35) = 0.1190
d) P(X > 31) = 0.8810
e) At least 35.7965 mpg
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation , the zscore of a measure X is given by:
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
a. P(X<30)
This is the pvalue of Z when X = 30. So
has a pvalue of 0.0392.
Then
P(X < 30) = 0.0392.
b) P(28 < X < 32)
This is the pvalue of Z when X = 32 subtracted by the pvalue of Z when X = 28. So
X = 32
has a pvalue of 0.2776.
X = 28
has a pvalue of 0.0016.
0.2776 - 0.0016 = 0.2760.
So
P(28 < X < 32) = 0.2760
c) P(X>35)
This is 1 subtracted by the pvalue of Z when X = 35. So
has a pvalue of 0.8810.
1 - 0.8810 = 0.1190
So
P(X > 35) = 0.1190
d. P(X>31)
This is 1 subtracted by the pvalue of Z when X = 31. So
has a pvalue of 0.1190.
1 - 0.1190 = 0.8810
So
P(X > 31) = 0.8810
e. the mileage rating that the upper 5% of cars achieve.
At least the 95th percentile.
The 95th percentile is X when Z has a pvalue of 0.95. So it is X when Z = 1.645. Then
At least 35.7965 mpg
painting for 20 minutes, took a shower
for 8 minutes, and read a book for 10
minutes. How many more minutes are
left before Kayla's bedtime?
Answer:
2 minutes are left
Step-by-step explanation:
1. Start with 40
2. Subtract 20 from 40 (40-20) to get 20 minutes.
3. Then take 8 minutes from 20 (20-8) to get 12 minutes
4. Lastly, take 10 minutes away from 12 (12-10) to get 2 minutes left.
5. This leaves Kayla with 2 minutes before bedtime.
By solving the equation "" we get "".
As we know the auxiliary equations,
here,
m = 1, 2, -2
Thus,
The general equation will be:
→
Particularly,
hence,
The complete equation will be:
→
Thus the above answer is right.
Learn more:
Answer:
Step-by-step explanation:
First of all write the auxialary equation as
m=1,2,-2
Hence general solution is
Particular solution of 5 is
Particular solution of e^x is
Particular solution of e^2x is
Together full solution is
702.96 sq ft
703.36 sq ft
703.96 sq ft
704.36 sq ft
Answer:
703.36 sq ft
Step-by-step explanation:
The lateral area is the product of the circumference and the height.
LA = 2π·r·h = 2·3.14·(7 ft)(16 ft) = 703.36 ft²
_____
Comment on the answer choices
Please note this is not the actual area, which rounds to 703.72 ft². The above value is obtained only by using an inappropriate value for π. The 5-significant-digit answer is not supported by a 3-significant-digit value for π. More appropriate would be π≈3.1416.