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2 years ago

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A researcher collated data on Americans’ leisure time activities. She found the mean number of hours spent watching television e

ach weekday to be 2. 7 hours with a standard deviation of 0. 4 hours. Jonathan believes that his football team buddies watch less television than the average American. He gathered data from 15 football teammates and found the mean to be 2. 3. Which of the following shows the correct z-statistic for this situation? -3. 87 -0. 26 3. 23 5. 22.

1 answer:

ikadub [295]2 years ago

6 0

The value of the z statistic for the considered data is given by: Option A: -3.87 approximately.

<h3>How to find the z score (z statistic) for the sample mean?</h3>

If we're given that:

Sample mean = $\overline{x}$
Sample size = n
Population mean = $\mu$
Sample standard deviation = s

Then, we get:

$z = \dfrac{\overline{x} - \mu}{s}$

If the sample standard deviation is not given, then we can estimate it(in some cases) by:

$s = \dfrac{\sigma}{\sqrt{n}}$

where $\sigma$ population standard deviation

For this case, we're specified that:

Sample mean = $\overline{x}$ = 2.3
Sample size = n = 15
Population mean = $\mu$ = 2.7
Population standard deviation = $\sigma$ = 0.4

Thus, the value of the z-statistic is evaluated as:

$z = \dfrac{\overline{x} - \mu}{\sigma/\sqrt{n}} = \dfrac{2.3- 2.7}{0.4/\sqrt{15}} \approx -3.87$

Thus, the value of the z statistic for the considered data is given by: Option A: -3.87 approximately.

Learn more about z statistic here:

brainly.com/question/27003351

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I plant grew 3 1/4 inches over 6 1/2 month period. What was the average monthly growth rate for the plant

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Answer:

0.5 inches per a month

Step-by-step explanation:

you turn them to a decimal to make them easier and then divide them to get the answer.

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3 years ago

The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as t

skad [1K]

Answer:

a) 0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

b) 0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

c) 0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

d) None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as the population mean and assume the population standard deviation of preparation fees is $100.

This means that $\mu = 273, \sigma = 100$

A) What is the probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 30, s = \frac{100}{\sqrt{30}}$

The probability is the p-value of Z when X = 273 + 16 = 289 subtracted by the p-value of Z when X = 273 - 16 = 257. So

X = 289

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{30}}}$

$Z = 0.88$

$Z = 0.88$ has a p-value of 0.8106

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{30}}}$

$Z = -0.88$

$Z = -0.88$ has a p-value of 0.1894

0.8106 - 0.1894 = 0.6212

0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

B) What is the probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 50, s = \frac{100}{\sqrt{50}}$

X = 289

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{50}}}$

$Z = 1.13$

$Z = 1.13$ has a p-value of 0.8708

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{50}}}$

$Z = -1.13$

$Z = -1.13$ has a p-value of 0.1292

0.8708 - 0.1292 = 0.7416

0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

C) What is the probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 100, s = \frac{100}{\sqrt{100}}$

X = 289

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{100}}}$

$Z = 1.6$

$Z = 1.6$ has a p-value of 0.9452

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{100}}}$

$Z = -1.6$

$Z = -1.6$ has a p-value of 0.0648

0.9452 - 0.0648 =

0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

D) Which, if any of the sample sizes in part (a), (b), and (c) would you recommend to ensure at least a .95 probability that the same mean is withing $16 of the population mean?

None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

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2 years ago

Need help with both questions please

Inessa [10]

9)To find the LCD you need to find a number that all of the denominator numbers can divide into, for this problem it would be 12. 12/6+12/4+12/3=1
2+3+4=1
9=1
No solution
10) In order to solve this problem you need to cross multiply
2(3x+5)=7(10
6x+10=70
6x=60
x=10
Have a great day!:)

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2 years ago

What transformation maps the preimage (red) onto the image (blue)?

Dominik [7]

Answer:

Rotate counter clockwise about the center point 180 degrees

Step-by-step explanation:

You can use transformations to preserve shape and size but move a shape or change its position. Transformations include:

reflection - a flip across an axis
rotation - a turn around a point
translation - a slide a specific distance and direction

To map the red into the blue, you can rotate it counter clockwise about the center point which the red and blue shapes share 180 degrees. It will land exactly on the red.

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3 years ago

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The average number of customers per day at a home improvement store is given in the following table. Let x represent the day whe

nika2105 [10]

Answer:

A. $33.690\sin (0.887x + 1.337) + 81.684$ .

Step-by-step explanation:

We are given that,

x = Number of days where 1 = Sun of 1st week and 7 = Sat of first week.

The corresponding table for the data is given by,

Days Day Number Number of Customers

Sun 1 115

Mon 2 77

Tue 3 60

Wed 4 51

Thur 5 68

Fri 6 86

Sat 7 120

The general form of the regression model is $a\sin (bx+c)+d$ .

Using the sinusoidal regression calculator, we get that,

a = 33.690

b = 0.887

c = 1.337

d = 81.684

That is, the sine regression model is $33.690\sin (0.887x + 1.337) + 81.684$ .

Thus, option A is the sine regression model for the given data.

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3 years ago

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