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

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

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 = (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 = (\sigma)/(√(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{(p(1-p))/(n)}`

In this question:

`p = 0.43, n = 165`

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

`s = \sqrt{(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 = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (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