# The following measurements were recorded for the drying time, in hours, of a certain brand of latex paint: 3.4, 2.5, 4.8, 2.9, 3.6, 2.8, 3.3, 5.6, 3.7, 2.8, 4.4, 4.0, 5.2, 3.0, 4.8. Assuming that the measurements represent a random sample from a normal population, find a 95% prediction interval for the drying time for the next trial of the paint.

The 95% confidence interval for the mean is (3.249, 4.324).

We can predict with 95% confidence that the next trial of the paint will be within 3.249 and 4.324.

Step-by-step explanation:

We have to calculate a 95% confidence interval for the mean.

As the population standard deviation is not known, we will use the sample standard deviation as an estimation.

The sample mean is:

The sample standard deviation is:

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=3.787.

The sample size is N=15.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

The t-value for a 95% confidence interval is t=2.145.

The margin of error (MOE) can be calculated as:

Then, the lower and upper bounds of the confidence interval are:

The 95% confidence interval for the mean is (3.249, 4.324).

## Related Questions

What is 6=- x/8
Plz help

1 .Multiply both sides by 88.

6\times 8=-x6×8=−x

2 .Simplify  6\times 86×8  to  4848.

48=-x48=−x

3. Multiply both sides by -1−1.

-48=x−48=x

4. x=−48

Step-by-step explanation:

If the level of significance of a hypothesis test is raised from .01 to .05, the probability of a Type II error a. will not change. b. will increase. c. will also increase from .01 to .05. d. will decrease.

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.

### Explanation:

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.

brainly.com/question/31665727

#SPJ3

What is the square root of 64y16?
o 44
O 4y
o 8
o gy

B is the answer of the question

Step-by-step explanation:

n+n= 2n

5(n+2)=5n+10

2(n+1)=2n+1

a 12-foot piece of string is cut into two pieces so that the longer piece is 3 feet longer than twice the shorter piece. find the length of both pieces. what is the length of the shorter piece

The longer piece would be 7.5f and the shorter 4.5

Which functions have a maximum value greater than the maximum of the function g(x) = –(x + 3)2 – 4? Check all that apply. f(x) = –(x + 1)2 – 2 f(x) = –|x + 4| – 5 f(x) = –|2x| + 3