Distribution of sample population

Question

Suppose 40% of the students in a university are baseball players.

If a sample of 743students is selected, what is the probability that the sample proportion of baseball players will differ from the population proportion by less than 3%? Round your answer to four decimal places.

Answer 1

Distribution of sample population

Suppose 40% of the students in a university are baseball players. If a sample of 743 students is selected, what is the probability that the sample proportion of baseball players will differ from the population proportion by less than 3%? Round your answer to four decimal places.

Given:

H₀ : p = 0.40

H₁ : p ≠ 0.40

p = population mean

p₁ = sample mean                                  

p = 0.40

p₁ = 0.37  or 0.43                              

n = 743

A student is either a baseball player or not a baseball player. This being a binary problem, we will apply binary probability distribution. Since:

np = 743 * 0.40 = 292.2 > 5

n(1-p) = 743 * 0.60 = 445.8 > 5

Hence, Central Limit Theorem (CLT) shall apply. That means, sample statistic will be very much like a normal distribution. In other words:

(p₁ - p)/ SQRT (p * (1 - p) /n) is approximately similar to N(0,1). Hence, we can use Standard Normal Distribution z-tables. Note that the denominator SQRT (p * (1 - p) /n) is the standard deviation of the binomial distribution.

Now, z = (p₁ – p) / SQRT (p * (1 - p) /n),  where z is z-statistic of the mean of the sampling distribution of the sample means (CLT).

z (0.37) = (0.37 – 0.40) /  SQRT(0.40 * 0.60/743) = -0.03 / 0.018 = -1.67

z (0.43) = (0.43 – 0.40) /  SQRT(0.40 * 0.60/743) = +0.03 / 0.018 = +1.67

The case is a two-tailed test where the level of significance is 3% or 0.03, with 0.015 on each tail.

From the excel, let us find out:

Zcrit (2-tailed) = -+ NORM.S(0.015) = -2.17 and +2.17

We will reject the NULL hypothesis if z < -2.17 & z> +2.17

But that is not the case. We can see:

z(0.37)  = -1.67 is greater than -2.17 and

z(0.43) = +1.67 is less than +2.17

 Hence, we fail to reject the NULL hypothesis. Further, on the right side probability that sample mean will differ by less than 1.5% is 98.5% and on the left side probability that sample mean will differ by less than 1.5% is 1.5%.

The acceptance zone of the NULL hypothesis is thus 97% or 0.97.

Answer: The probability that the sample proportion of baseball players will differ from the population proportion by less than 3% is 0.97.