Question

The University of Arkansas recently reported that 43% of college students aged 18-24 would spend their spring break relaxing at home. A sample of 165 college students is selected.
a. Calculate the appropriate standard error calculation for the data.
b. What is probability that more than 50% of the college students from the sample spent their spring breaks relaxing at home?

Answer 1

Answer:

a. 0.0385

b. 3.44% probability that more than 50% of the college students from the sample spent their spring breaks relaxing at home

Step-by-step explanation:

To solve this question, we need to understand 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.

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

In this question:

$p = 0.43, n = 165$

a. Calculate the appropriate standard error calculation for the data.

$s = \sqrt{\frac{0.43*0.57}{165}} = 0.0385$

b. What is probability that more than 50% of the college students from the sample spent their spring breaks relaxing at home?

This is 1 subtracted by the pvalue of Z when X = 0.5. So

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

By the Central Limit Theorem

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

$Z = \frac{0.5 - 0.43}{0.0385}$

$Z = 1.82$

$Z = 1.82$ has a pvalue of 0.9656

1 - 0.9656 = 0.0344

3.44% probability that more than 50% of the college students from the sample spent their spring breaks relaxing at home