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Juliette [100K]
4 years ago
9

Suppose that a recent poll found that 50 ​% of adults in a certain country believe that the overall state of moral values is poo

r. If a survey of a random sample of 15 adults in this country is conducted in which they are asked to disclose their feelings on the overall state of moral​ values, complete parts​ (a) through​ (e) below.
Mathematics
1 answer:
Bumek [7]4 years ago
4 0

Answer: 3.75

Step-by-step explanation:

P=0.50, n=15

np= 0.5x15 gives 7.5

n(1-p)= 15x(1-0.5)=7.5

Both are greater than 5

Then Mean, u=np;15x0.5=7.5

Standard deviation, √np(1-p)

15x0.5(0.5)= 3.75

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She has totally 10+5+3=18 marbles, so the probability is 10/18=5/9. The complement is (9-5)/9=4/9.

Hope this helped!
4 0
3 years ago
In 1990, it cost $800,000 for a 30-second commercial during the Super Bowl. In 2020, the cost for a 30-second commercial was $5,
lesya [120]

Answer:

690%

Step-by-step explanation:

To find the percent increase, take the new amount and subtract the original amount

5600000-80000=5520000

Divide it by the original amount

5520000/800000

6.9

Multiply this by 100% to change from decimal form to percent form

6.9*100%

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3 years ago
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Your question but don't know how to solve it like a dog

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In his free time, Gary spends 13 hours per week on the Internet and 13 hours per week playing video games. If Gary has five hour
goldenfox [79]
<span>Gary spend 13 hours per week on the Internet and 13 hours on video games Gary has 5 hours of free time each day,
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26/35 (that is a fraction) x100 =</span><span>74.285714 
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6 0
3 years ago
Sample Size for Proportion As a manufacturer of golf equipment, the Spalding Corporation wants to estimate the proportion of gol
Dima020 [189]

Answer:

n=\frac{0.5(1-0.5)}{(\frac{0.025}{2.58})^2}=2662.56  

And rounded up we have that n=2663

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

\hat p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. And the critical value would be given by:

t_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.025 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

We can assume an estimated proportion of \hat p =0.5 since we don't have prior info provided. And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.025}{2.58})^2}=2662.56  

And rounded up we have that n=2663

6 0
4 years ago
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