Question

At a university, 60% of the 7,400 students are female. The student newspaper reports the results of a survey of a random sample of 50 students about various topics involving alcohol abuse. They report that their sample contained 26 females. Let X be a random variable such that x = 1 denotes female and x = 0 denotes male. Consider the sampling distribution of the sample proportion of females in samples of size 50. State the mean and standard error for this sampling distribution.

Answer 1

Given Information:

Population mean = p  = 60% = 0.60

Population size = N = 7400

Sample size = n = 50

Required Information:

Sample mean = μ = ?

standard deviation = σ = ?

Answer:

Sample mean = μ = 0.60

standard deviation = σ = 0.069

Step-by-step explanation:

We know from the central limit theorem, the sampling distribution is approximately normal as long as the expected number of successes and failures are equal or greater than 10

np ≥ 10

50*0.60 ≥ 10

30 ≥ 10 (satisfied)

n(1 - p) ≥ 10

50(1 - 0.60) ≥ 10

50(0.40) ≥ 10

20 ≥ 10  (satisfied)

The mean of the sampling distribution will be same as population mean that is

Sample mean = p = μ = 0.60

The standard deviation for this sampling distribution is given by

$\sigma = \sqrt{\frac{p(1-p)}{n} }$

Where p is the population mean that is proportion of female students and n is the sample size.

$\sigma = \sqrt{\frac{0.60(1-0.60)}{50} }\\\\\sigma = \sqrt{\frac{0.60(0.40)}{50} }\\\\\sigma = \sqrt{\frac{0.24}{50} }\\\\\sigma = \sqrt{0.0048} }\\\\\sigma = 0.069$

Therefore, the standard deviation of the sampling distribution is 0.069.