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bezimeni [28]
2 years ago
12

a senator wishes to estimate the proportion of united states voters who favor new road construction. what size sample should be

obtained in order to be 95% confident
Mathematics
1 answer:
Nitella [24]2 years ago
8 0

The  size of the sample should be obtained at 2401 in order to be 95% confident

Given that:

Margi of error, ME=0.02

Population proportion, p=0.50

To Find: The size of the sample that should be obtained in order to be 95% confident

Let the size of the sample be n

Using formula,

n=(p*(1-p))*(z(0.05/2)/ME)^{2} \\n=(0.5*(1-0.5))*(1.96/0.02)^{2} \\n=0.25*9604\\n=2401

Therefore, the size of the sample should be obtained at 2401 in order to be 95% confident

Learn more about Population proportion here brainly.com/question/15703406

#SPJ4

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2 1/2 x 3 1/5 fast please
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Step-by-step explanation:

<u>Let's turn each mixed fraction into an improper fraction:</u>

2 \frac{1}{2} =\frac{2}{2} +\frac{2}{2} +\frac{1}{2} =\frac{5}{2}

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There are 16 girls and 18 boys in a class. The teacher Chooses a students name at random to answer a question. What is the proba
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A certain car model has a mean gas mileage of 34 miles per gallon (mpg) with a standard deviation A pizza delivery company buys
zubka84 [21]

Answer:

z =\frac{33.3- 34}{\frac{5}{\sqrt{54}}}= -1.028

z =\frac{34.3- 34}{\frac{5}{\sqrt{54}}}= 0.441

An we can use the normal standard table and the following difference and we got this result:

P(-1.028

Step-by-step explanation:

Assuming this statement to complete the problem "with a standard deviation 5 mpg"

We have the following info given:

\mu = 34 represent the mean

\sigma= 5 represent the deviation

We have a sample size of n = 54 and we want to find this probability:

P(33.3 < \bar X< 34.3)

And for this case since the sample size is large enough >30 we can apply the central limit theorem and then we can use this distribution:

\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

And we can use the z score formula given by:

z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

z =\frac{33.3- 34}{\frac{5}{\sqrt{54}}}= -1.028

z =\frac{34.3- 34}{\frac{5}{\sqrt{54}}}= 0.441

An we can use the normal standard table and the following difference and we got this result:

P(-1.028

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