Question

A researcher reports survey results by stating that the standard error of the mean is 25 the population standard deviation is 400. Use z-table.

a. How large was the sample used in this survey?
b. What is the probability that the point estimate was within ±15 of the population mean?

Answer 1

Answer:

a) A sample of 256 was used in this survey.

b) 45.14% probability that the point estimate was within ±15 of the population mean

Step-by-step explanation:

This question is solved using the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore 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 pvalue, 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 $\mu$ 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.

a. How large was the sample used in this survey?

We have that $s = 25, \sigma = 400$. We want to find n, so:

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

$25 = \frac{400}{\sqrt{n}}$

$25\sqrt{n} = 400$

$\sqrt{n} = \frac{400}{25}$

$\sqrt{n} = 16$

$(\sqrt{n})^2 = 16^2[tex]

[tex]n = 256$

A sample of 256 was used in this survey.

b. What is the probability that the point estimate was within ±15 of the population mean?

15 is the bounds with want, 25 is the standard error. So

Z = 15/25 = 0.6 has a pvalue of 0.7257

Z = -15/25 = -0.6 has a pvalue of 0.2743

0.7257 - 0.2743 = 0.4514

45.14% probability that the point estimate was within ±15 of the population mean