Mathematics, 19.05.2021 18:10 haileybug0305

An important aspect of a federal economic plan was that consumers would save a substantial portion of the money that they received from an income tax reduction. Suppose that early estimates of the portion of total tax saved, based on a random sampling of 35 economists, had mean 26% and standard deviation 12%. Required:
What is the approximate probability that a sample mean estimate, based on a random sample of n = 35 economists, will lie within 1% of the mean of the population of the estimates of all economists?

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0.3758 = 37.58% probability that a sample mean estimate will lie within 1% of the mean of the population of the estimates of all economists.

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.

Mean 26% and standard deviation 12%.

This means that $\mu = 26, \sigma = 12$

Sample of 35:

This means that $n = 35, s = \frac{12}{\sqrt{35}}$

What is the approximate probability that a sample mean estimate, based on a random sample of n = 35 economists, will lie within 1% of the mean of the population of the estimates of all economists?

This is the p-value of Z when X = 26 + 1 = 27 subtracted by the p-value of Z when X = 26 - 1 = 25. So

X = 27

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

By the Central Limit Theorem

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

$Z = \frac{27 - 26}{\frac{12}{\sqrt{35}}}$

Z = 0.49

Z = 0.49 has a p-value of 0.6879

X = 25

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

$Z = \frac{25 - 26}{\frac{12}{\sqrt{35}}}$

Z = -0.49

Z = -0.49 has a p-value of 0.3121

0.6879 - 0.3121 = 0.3758

0.3758 = 37.58% probability that a sample mean estimate will lie within 1% of the mean of the population of the estimates of all economists.

Answer from: Quest

y - y1 = m(x- x1)

y - (-5) = 2/3(x -0)

y + 5 = 2/3x

y = 2/3x - 5

Answer from: Quest

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step-by-step explanation:

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