In the checksums, you validated that the average SAT score for all of the records is 1,059.07. When we work with the data more rigorously, several tests will require us to transform NULL values. If you were to transform the NULL SAT values into 0, what would happen to the average (would it stay the same, decrease, or increase?

Answers

Answer 1

Answer:

Answer:

It would decrease.

Step-by-step explanation:

The null almost always acts like a placeholder where there is nothing to record for that entry, we can say that it is a dummy character.

The difference between null and zero is that zero has a value in a context whereas null does not have any value.

So given the example in the question, if we replace the null values with zeros, that would affect the average because of the fact that zero values are going to get included in the average, null values were not included. Therefore this will result in the average being decreased.

I hope this answer helps.

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Factor 6q2 − 41q + 30

Answers

(X-6)(6x-5)
So;
X-6=0
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People tend to evaluate the quality of their lives relative to others around them (Frieswijk et al., 2004). In one study, researchers conducted interviews with n = 9 frail elderly people. During the interview, each person was compared with a fictitious person who was worse off than the elderly person. The scores below are the measures from a life-satisfaction scale for the elderly sample. Assume that the average score on this scale in the population is u = 20. Are the data sufficient to conclude that the elderly people in this sample are either significantly more or less satisfied than others in the general population? The life-satisfaction scores for the sample are 18, 23, 24, 22, 19, 27, 23, 26, 25. a. Which kind of t-test should you use? b. How many tails should the test have? Circle a word or phrase in the problem that told you this. c. State the null and alternative hypotheses in statistical notation: d. Determine the critical t using an alpha = .05. Sketch the null distribution, note the location of the critical t, and shade the critical region. e. Calculate the t-statistic and plot it on the sketch you drew above. f. Make a decision (either reject the null or fail to reject it)

Answers

Answer:

Step-by-step explanation:

Hello!

Research.

n=9 frail elderly were interview and compared to a fictitious person who was worse off then the interviewee, a life-satisfaction score was determined for each person.

18, 23, 24, 22, 19, 27, 23, 26, 25

Assuming that the population average score is μ= 20, the researchers think that the elderly in the sample are more or less satisfied than others in the general population.

a. You have the information of one sample, assuming this sample has a normal distribution and each elderly interviewed is independent, then the t-test of choice is a one-sample t-test.

b. and c. If you say that the elderly are "more or less" satisfied than the others, this means that they are either as satisfied as to the general population or not satisfied as to the general population. Symbolically:

H₀: μ = 20

H₁: μ ≠ 20

This is a two-tailed test, meaning, you will have two critical regions.

α: 0.05

Left critical value: $t_(n-1;/\alpha 2) = t_(8; 0.025)= -2.306$

Right critical value: $t_(n-1;1-\alpha /2) = t_(8;0.975) = 2.306$

$t_(H_0)= (X[bar]-Mu)/((S)/(√(n) ) ) ~t_(n-1)$

X[bar]= 23

S= 3

$t_(H_0)= (23-20)/((3)/(√(9) ) )= 3$

Considering that the calculated t-value is greater than the right critical value, the decision is to reject the null hypothesis, so using a significance level of 5% you can conclude that the average life-satisfaction score of the elderly is different than 20.

I hope it helps!

A certain firm has plants a, b, and c producing respectively 35\%, 15\%, and 50\% of the total output. The probabilities of a nondefective product are, respectively, 0.75, 0.95, and 0.85. A customer receives a defective product. What is the probability that it came from plant c?

Answers

The proportion of production that is defective and from plant A is

... 0.35·0.25 = 0.0875

The proportion of production that is defective and from plant B is

... 0.15·0.05 = 0.0075

The proportion of production that is defective and from plant C is

... 0.50·0.15 = 0.075

Thus, the proportion of defective product that is from plant C is

... 0.075/(0.0875 +0.0075 +0.075) = 75/170 = 15/34 ≈ 44.12%

_____

P(C | defective) = P(C&defective)/P(defective)

Final answer:

The question required the use of Bayes' theorem to determine the probability of a defective product coming from plant c. Given the probabilities of defectiveness for each plant, the calculation indicated that there is approximately a 54.55% chance that a defective product came from plant c.

Explanation:

The problem described can be solved using Bayes' theorem, which is a principle in Probability that is used when we need to revise/or update the probabilities of events given new data. Since a defective product is received, and we need to determine the probability of it coming from plant c, we apply Bayes' theorem for the probability of events a, b, and c (representative of the products from the respective plants).

The Bayesian formula we will use, given the probabilities of a, b and c respectively and the probability of receiving a nondefective product from these plants, is: P(c|defective) = [P(defective|c) * P(c)] / [P(defective|a) * P(a) + P(defective|b) * P(b) + P(defective|c) * P(c)].

First, calculate the probability of a defective product from each plant (1 minus the probability of a nondefective product): these are 0.25 for plant a, 0.05 for plant b, and 0.15 for plant c.

Then substitute the values: P(c|defective) = [0.15 * 0.50] / [(0.25 * 0.35) + (0.05 * 0.15) + (0.15 * 0.5)] = 0.075 / 0.1375 = 0.5454545.

So, given a defective product, there is approximately a 54.55% chance that it was produced by plant c.

Learn more about Bayes' theorem here:

brainly.com/question/34293532

#SPJ11

Find the distance between the point (-18,-19) and the point (-18,16).

Answers

ANSWER

EXPLANATION

To find the distance between the two points, we will use the formula:

$\begin{gathered} D=\text{ }\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \n \text{where (x}_1,y_1)\text{ = (-18, -19)} \n (x_2,y_2)\text{ = (-18, 16)} \end{gathered}$

So, we have that:

$undefined$

Solve m^2+8m=-3 by completing the square

Answers

i hope this helps you!!! be safe

Find the slope Y=5x+12

Answers

Answer: The slope = 5.

Step-by-step explanation:

The equation of a line in the slope -intercept form is given by :-

$y=mx+c$

,where m= slope , c= y-intercept.

The slope of a line represents the rate of change of y with respect to x.

Given: Equation of line : Y= 5x+12

Here Slope (m)=5

Hence, the slope = 5.

Hi!

$\spadesuit$ We're given an equation in slope-intercept form:

$\clubsuit~~~\Large\boldsymbol{y=mx+b} ~~~\clubsuit$

Where:

$\diamondsuit~~~\large\boldsymbol{m- > slope}$

$\diamondsuit~~\large\boldsymbol{b- > y-intercept}$

Notes:

Please remember that the slope is the coefficient of x (the number that comes just before x; in 5x, the coefficient of x is 5), and the y-intercept is added to or subtracted from the slope.

Therefore,

$\stackrel{\heartsuit}{\boxed{\boxed{\textsc{Answers:}\begin{cases}\bold{Slope:5}\n\bold{12}\end{cases}}}}$

Cheers!!

I hope that the explanation helped you out! :D

If you need any explanation or clarification, please let me know!

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A dealer advertises that a certain brand of tire averages 50,000 miles of use before needing to be replaced. Testing this claim, an investigator finds that a sample of 19 randomly selected tires from this dealer had an average lifespan of 48,700 miles with a standard deviation of 4,500 miles. Assume tire life spans are approximately normally distributed. At the α = 0.05 level, find if the investigator can prove the dealer is exaggerating the average lifespan of their tires by performing the hypothesis test.

The table shows the price for different numbers of notebooks:Number of Notebooks Price (in dollars)2 123 184 5 30

Solve for the value of c.

Answer and I will give you brainiliest