Suppose that there is a school bond referendum in Greensboro, and 73% of voters support it. You randomly ask 20 Greensboro voters whether they support the bond referendum. The standard error of the sample proportion is _____. The probability that 12 or fewer people (out of 20) in your sample support the bond referendum is _____. Group of answer choices 0.0993; 0.6000 0.0993; 0.0951 0.1095; 0.1170 0.1095; 0.8830

Answer:

The standard error of the sample proportion is 0.0993.

The probability that 12 or fewer people (out of 20) in your sample support the bond referendum is 0.0951.

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

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

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

73% of voters support it.

This means that [tex]p = 0.73[/tex]

Sample of 20 voters

This means that [tex]n = 20[/tex]

Mean and standard deviation:

[tex]\mu = p = 0.73[/tex]

[tex]s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.73*0.27}{20}} = 0.0993[/tex]

The standard error of the sample proportion is 0.0993.

The probability that 12 or fewer people (out of 20) in your sample support the bond referendum is

12/20 = 0.6, so this is the p-value of Z when X = 0.6.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{0.6 - 0.73}{0.0993}[/tex]

[tex]Z = -1.31[/tex]

[tex]Z = -1.31[/tex] has a p-value of 0.0951.

So

The probability that 12 or fewer people (out of 20) in your sample support the bond referendum is 0.0951.

The standard error of the sample proportion is 0.0098.

The probability that 12 or fewer people (out of 20) in your sample support the bond referendum is 0.0951

Given that,

Suppose that there is a school bond referendum in Greensboro, and 73% of voters support it.

You randomly ask 20 Greensboro voters whether they support the bond referendum.

We have to determine,

The standard error of the sample proportion is.

The probability that 12 or fewer people (out of 20) in your sample support the bond referendum is .

According to the question,

There is a school bond referendum in Greensboro, and 73% of voters support it.

20 Greensboro voters whether they support the bond referendum.

[tex]standard \ deviation = \sqrt{\dfrac{p(1-p)}{n}}[/tex]

Where, p = 73% = 0.73

And n = 20

Therefore,

[tex]Standard \ deviation = \sqrt{\dfrac{p(1-p)}{n}}\\\\ Standard \ deviation = \sqrt{\dfrac{0.73(1-0.73)}{20}}\\\\Standard \ deviation = \sqrt{\dfrac{0.73\times 0.27}{20}}\\\\Standard \ deviation = \sqrt{\dfrac{0.19}{20}}\\\\Standard \ deviation = 0.0098[/tex]

The standard error of the sample proportion is 0.0098.

[tex]\dfrac{12}{20 }= 0.6[/tex]

 This is the p-value of Z when X = 0.6.

Therefore,

[tex]Z = \dfrac{X-\mu}{\sigma}\\\\Z = \dfrac{0.6-0.73}{0.993}\\\\Z = -1.31\\\\[/tex]

Z = -1.31 has a p-value of 0.0951.

Hence, The probability that 12 or fewer people (out of 20) in your sample support the bond referendum is 0.0951.

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