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kaheart [24]
1 year ago
12

Which of the following statements are true concerning the mean of the differences between two dependent samples​ (matched pairs)

? Select all that apply. A. The methods used to evaluate the mean of the differences between two dependent variables apply if one has 86 IQ scores of taxpayers from Texas and 86 IQ scores of taxpayers from Ohio. B. The requirement of a simple random sample is satisfied if we have matched pairs of voluntary response data. C. If one has more than 23 matched pairs of sample​ data, one can consider the sample to be large and there is no need to check for normality. D. If one has twenty matched pairs of sample​ data, there is a loose requirement that the twenty differences appear to be from a normally distributed population. E. If one wants to use a confidence interval to test the claim that mu Subscript d Baseline greater than 0 with a 0.01 significance​ level, the confidence interval should have a confidence level of 98​%.
Mathematics
1 answer:
elena-s [515]1 year ago
7 0

Answer:

A and C

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations we can use it.  

And in order to conduct it we need to have some assumptions:

1) The dependent variable must be continuous (interval/ratio).

2) The observations are independent of one another.

3) The dependent variable should be approximately normally distributed.

4) The dependent variable should not contain any outliers.

Let's analyze one by one the options on this case:

A. The methods used to evaluate the mean of the differences between two dependent variables apply if one has 86 IQ scores of taxpayers from Texas and 86 IQ scores of taxpayers from Ohio

We can assume that is true since the two variables are dependent and we are assuming that all the other conditions are satisfied to use the t paired test.

B. The requirement of a simple random sample is satisfied if we have matched pairs of voluntary response data.

That's not neccesary true since is not a requirement in order to use the t pairedtest.

C. If one has more than 23 matched pairs of sample​ data, one can consider the sample to be large and there is no need to check for normality.

We can assume that is true since we need to ensure normality in order to apply the test and if the sample size is large enough large we can apply the test.

D. If one has twenty matched pairs of sample​ data, there is a loose requirement that the twenty differences appear to be from a normally distributed population.

False the requirement of normality is important to apply the test not a loose requirement.

E. If one wants to use a confidence interval to test the claim that mu Subscript d Baseline greater than 0 with a 0.01 significance​ level, the confidence interval should have a confidence level of 98​%.

That's not correct the significance level is \alpha=1-0.98=0.02 and that not correspond to the significance level given on this case.

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A cement bridge post is 24 inches square and 15 feet 9 inches in length. If the cement weighs 145 pounds per cubic foot, how muc
lisov135 [29]

Answer:

The answer is option (C), One cement bridge post weighs 9,135 pounds

Step-by-step explanation:

Step 1: Get the total volume of a cement bridge post

The cement bridge post is in the shape of a cuboid, therefor the volume of the cement bridge post can be expressed as;

Volume of cement bridge post=Base area×Length

where;

Base area=(24^2) inches square

1 foot=12 inches

Convert 24 inches to foot=24/12=2^2=4 feet²

Length=15 feet and 9 inches

1 foot=12 inches

Convert 9 inches to foot=9/12=0.75 feet

Total length=(15+0.75)=15.75 feet

replacing;

Volume of cement bridge post=(4×15.75)=63 cubic feet

Volume of cement bridge post=63 cubic feet

Step 2: Get the total weight of the cement bridge post

Total weight of the cement bridge post=Weight per cubic foot×total volume of the cement bridge post

where;

Weight per cubic foot=145 pounds per cubic foot

Total volume of the cement bridge post=63 cubic feet

replacing;

Total weight of the cement bridge post=(145×63)=9,135 pounds

One cement bridge post weighs 9,135 pounds

8 0
1 year ago
Meeting at least one person with the flu in thirteen random encounters on campus when the infection rate is 2% (2 in 100 people
s344n2d4d5 [400]

Answer:

23.1% probability of meeting at least one person with the flu

Step-by-step explanation:

For each encounter, there are only two possible outcomes. Either the person has the flu, or the person does not. The probability of a person having the flu is independent of any other person. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Infection rate of 2%

This means that p = 0.02

Thirteen random encounters

This means that n = 13

Probability of meeting at least one person with the flu

Either you meet none, or you meet at least one. The sum of the probabilities of these outcomes is 1. So

P(X = 0) + P(X \geq 1) = 1

We want P(X \geq 1). Then

P(X \geq 1) = 1 - P(X = 0)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{13,0}.(0.02)^{0}.(0.98)^{13} = 0.7690

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.769 = 0.231

23.1% probability of meeting at least one person with the flu

6 0
1 year ago
What single decimal multiplier would you use to increase by 15% followed by a 14% increase?
Yuliya22 [10]

Answer:

The single decimal multiplier that I will use to increase by 15% followed by a 14% increase is 1.311

Step-by-step explanation:

Let

x ----> the number

Remember that

100\%+15\%=115\%=115/100=1.15

100\%+14\%=114\%=114/100=1.14

we know that

Increase a number by 15% followed by a 14% increase is the same that multiply the number for (1.15)(1.14)

so

x(1.15)(1.14)=1.311x

therefore

The single decimal multiplier that I will use to increase by 15% followed by a 14% increase is 1.311

3 0
1 year ago
Zachary weighs 37 kilograms 95 grams. Gabe weighs 4,650 grams less than Zachary. Harry weighs 2,905 grams less than Gabe. How mu
Archy [21]
First make the numbers the same.
37 kilograms is 37000 grams, for there are 1000 grams in a kilograms.
Zachary weighs 37095 grams
37095-4650= Gabes weight = 32445 grams
32445-2905= Harry's weight = 29540 grams or 29,54 kilograms
5 0
2 years ago
If speed varies inversely as the time it takes to drive and Kris takes 5 hours driving at 55 mph, what speed will Martin need to
Fudgin [204]

Answer:

<em>55mph . None of the options are correct </em>

Step-by-step explanation:

If the speed varies inversely as the time it takes to drive, then v ∝ 1/t. where;

v is the speed

t is the time taken

Hence;

v = k/t with k being the constant of proportionality.

Since it takes Kris 5 hours when driving at 55 mph, we will substitute v = 55mph and t = 5 hours. into the equation above to get the value of k as shown:

55 = k/5

Cross multiply

k = 55*5

k = 275

Hence, to calculate the speed it will Martin to  drive for 5 hours, we will substitute k = 275 and t = 5 into the original equation v = k/t  as shownl

v = 275/5

v = 55 mph

<em>Hence, we can conclude that Martin will also need to drive at a speed of 55mph if he wants to take 5hours.</em>

3 0
2 years ago
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