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
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
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
. We want to find n, so:
![s = \frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
![25 = \frac{400}{\sqrt{n}}](https://tex.z-dn.net/?f=25%20%3D%20%5Cfrac%7B400%7D%7B%5Csqrt%7Bn%7D%7D)
![25\sqrt{n} = 400](https://tex.z-dn.net/?f=25%5Csqrt%7Bn%7D%20%3D%20400)
![\sqrt{n} = \frac{400}{25}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B400%7D%7B25%7D)
![\sqrt{n} = 16](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%2016)
![(\sqrt{n})^2 = 16^2[tex][tex]n = 256](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E2%20%3D%2016%5E2%5Btex%5D%3C%2Fp%3E%3Cp%3E%5Btex%5Dn%20%3D%20256)
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