Question

An investment website can tell what devices are used to access the site. The site managers wonder whether they should enhance the facilities for trading via​ "smart phones", so they want to estimate the proportion of users who access the site that way​(even if they also use their computers​ sometimes). They draw a random sample of 100 investors from their customers. Suppose that the true proportion of smart phone users is 33​%.

a) What would the standard deviation of the sampling distribution of the proportion of the smart phone users​ be?
b) What is the probability that the sample proportion of smart phone users is greater than
0.33?
c) What is the probability that the sample proportion is between 0.19 and 0.31​?

Answer 1

a) 0.047

b) 50% probability that the sample proportion of smart phone users is greater than 0.33.

c) 33.39% probability that the sample proportion is between 0.19 and 0.31

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, we have that:

`p = 0.33, n = 100`

a) What would the standard deviation of the sampling distribution of the proportion of the smart phone users​ be?

`s = \sqrt{(0.33*0.67)/(100)} = 0.047`

b) What is the probability that the sample proportion of smart phone users is greater than 0.33?

This is 1 subtracted by the pvalue of Z when X = 0.33. So

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

By the Central Limit Theorem

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

`Z = (0.33 - 0.33)/(0.047)`

`Z = 0`

`Z = 0` has a pvalue of 0.5

1 - 0.5 = 0.5

50% probability that the sample proportion of smart phone users is greater than 0.33.

c) What is the probability that the sample proportion is between 0.19 and 0.31​?

This is the pvalue of Z when X = 0.31 subtracted by the pvalue of Z when X = 0.19. So

X = 0.31

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

`Z = (0.31 - 0.33)/(0.047)`

`Z = -0.425`

`Z = -0.425` has a pvalue of 0.3354

X = 0.19

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

`Z = (0.19 - 0.33)/(0.047)`

`Z = -2.97`

`Z = -2.97` has a pvalue of 0.0015

0.3354 - 0.0015 = 0.3339

33.39% probability that the sample proportion is between 0.19 and 0.31