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makvit
2 months ago
13

A statistics practitioner in a large university is investigating the factors that affect salary of professors. He wondered if ev

aluations by students are related to salaries. To this end, he collected 100 observations on:
y = Annual salary (in dollars)
x = Mean score on teaching evaluation

To accomplish his goal, he assumes the following relationship:

y = β(0) + β(1)x + ε

Then, using Excel’s Data Analysis, he obtained the following result.

R2=0.23

Coefficient Standard Error
Intercept 25675.5 11393
x 5321 2119

What is the p-value?

Select one:

A. 0.02
B. 0.01
C. 0.17
D. 0.34
Mathematics
1 answer:
tester [12.3K]2 months ago
3 0
In this context, we analyze a linear regression where Y is the variable "Annual salary" predicted by X, which denotes "Mean score on teaching evaluation" for university professors. The goal is to ascertain whether there is a correlation between student evaluations and professor salaries. The population regression equation can be stated as E(Y)= β₀ + β₁X. Given an n = 100 sample, data shows an R² of 0.23. Further statistical calculations yield the estimated equation as ^Y= 25675.5 + 5321X. To verify if teaching evaluations impact salaries, the null and alternative hypotheses are H₀: β = 0 and H₁: β ≠ 0. A two-tailed t-test can be performed, with the calculated t-value being approximately 25.1109. The resulting two-tailed p-value is found to be significantly less than 0.00001.
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Suppose your statistics professor reports test grades as​ z-scores, and you got a score of 1.57 on an exam. ​a) Write a sentence
tester [12383]

Answer:

a) In this case, we have a z-score of 1.57, characterized as:

z =\frac{x -\mu}{\sigma}

This signifies that our score is 1.57 standard deviations above the average of all test scores.

b) P(Z

Using the normal standard distribution or Excel, we computed:

P(Z

This represents 2.275% of the dataset.

Step-by-step explanation:

Previous concepts

Normal distribution denotes a "probability distribution that is symmetric around the mean, indicating that data close to the mean occurs more frequently than data further from it".

The Z-score serves as "a statistical measurement representing a value's relation to the average (mean) of a set, calculated in terms of standard deviations away from the mean".

Solution to the problem

Part a

For this instance, we hold a z-score of 1.57, which is defined as:

z =\frac{x -\mu}{\sigma}

This shows that our score is 1.57 deviations above the overall test score average.

Part b

A z-score of z=-2 indicates that your friend's score is 2 deviations below the other test scores.

Assuming a normal distribution, we can derive the percentage:

P(Z

Using the normal standard distribution or Excel, we discovered:

P(Z

This represents 2.275% of the data.

6 0
3 months ago
A gardener has a cylindrical planter that has a diameter of 16 inches and is 2 feet deep. What is the approximate number of cubi
tester [12383]
The response is 16.75 ft3
4 0
1 month ago
The weights (to the nearest pound) of some boxes to be shipped are found to be:. Weight 65 68 69 70 71 72 90 95 frequency 1 2 5
zzz [12365]
Thus, the average is 72.97

We must deduct the mean from each number and square the result.
(65-72.97)^2= 63.5209
(68-72.97)^2=24.7009
(69-72.97)^2=15.7609
(70-72.97)^2=8.8209
(71-72.97)^2= 3.8809
(72-72.97)^2=0.9409
(90-72.97)^2=290.0209
(95-72.97)^2=485.3209

Next, we sum these new values (also taking their frequencies into account) and compute their average.
Sum the figures
63.5209+(2 •24.7009=49.4018)+(5•15.7609=78.8045)+(8•8.8209=70.5672)+(7•3.8809=27.1663)+(3•0.9409=2.8227)+(2•290.0209=580.0418)+(2•485.3209=970.6418)= 1,842.967
Then divide by the total number to determine the mean
1,842.967/ 30=61.43223333

The standard deviation is the square root of the average which equals
Square root of 61.43223333=7.837871735
Round it to the nearest tenth
The Standard Deviation is 7.8



7 0
2 months ago
A college student is interested in testing whether business majors or liberal arts majors are better at trivia. The student give
Leona [12618]

Answer:

There is no notable difference between the two averages at a 5% significance level.

Step-by-step explanation:

A college student wanted to investigate whether business majors or liberal arts majors performed better in trivia.

The student administered a trivia quiz to 30 business majors, achieving an average score of 86. The same quiz was given to 30 liberal arts majors, with an average score of 89.

This process involved hypothesis testing to compare the means of two different disciplines. n =30

H_0: Mean of business majors = Mean of liberal arts majos\\H_a:Mean of business majors \neq Mean of liberal arts majos

Since no specific claim is made regarding which group is better, a two-tailed test is appropriate.

The p-value achieved = 0.0524 >5% exceeds our alpha level,

indicating no significant difference exists between the average scores of the two groups, so we accept the null hypothesis.

4 0
3 months ago
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