Question

Question 3​

It is believed that 35% of the population approves the current government. A poll is conducted prior to elections.

a) Find the mean and standard error of the sample proportion, , of people approving the government.

b) What is the sampling distribution of ? State any assumptions you need to make.

c) A random sample of people were asked if they approve the government. What is the probability that at least 150 out of the 400 replied ‘yes’, while the remaining said ‘no’?

d) What is the probability that if a different similar random sample is taken, it will have a sample proportion less than 0.32?

Answer 1

Solution Q 3

Back-up Theory

If p = population proportion and p_hat = sample proportion based on a sample of size n, then

Mean of p_hat = p .............................................................................................................................. (1)

Standard deviation of p_hat = √{p(1 - p)/n}.......................................................................................... (2)

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where

n = number of trials and p = probability of one success, then,

probability mass function (pmf) of X is given by p(x) = P(X = x) = (ⁿC_x)(p^x)(1 - p)^{n – x}.......…….…..(3)

[This probability can also be directly obtained using Excel Function: Statistical, BINOMDIST] ..….(3a)

If X ~ B(n, p), np ≥ 10 and np(1 - p) ≥ 10, then Binomial probability can be approximated by

Standard Normal probabilities by Z = (X – np)/√{np(1 - p)} ~ N(0, 1) …....................……....………..(4)

Or, equivalently, Z = (p_hat – p)/√{p(1 - p)/p} ~ N(0, 1) ………………........................................……..(4a)

If a random variable X ~ N(µ, σ²), i.e., X has Normal Distribution with mean µ and variance σ²,

then, Z = (X - µ)/σ ~ N(0, 1), i.e., Standard Normal Distribution and hence

P(X ≤ or ≥ t) = P[{(X - µ)/σ} ≤ or ≥ {(t - µ)/σ}] = P[Z ≤ or ≥ {(t - µ)/σ}] .…………….....................…...…(5)

Probability values for N(0, 1) can be directly read off from Standard Normal Tables….......……...... (5a)

or can be found using Excel Function: Statistical, NORMSDIST(z) which gives P(Z ≤ z) ….............(5b)

Now to work out the solution,

In the given scenario, p = 0.35 [i.e., 35%] ........................................................................................ (6)

Part (a)

Vide (1) and (6),

Mean of the sample proportion of people approving the government = 0.35 Answer 1

Vide (2) and (6),

Standard error of the sample proportion of people approving the government

= √{0.35 x 0.65/n}

= √{0.35 x 0.65/400}, taking n = 400

= 0.0238 Answer 2

Part (b)

Vide (4a), sampling distribution of sample proportion is: N(0, 1)   Answer 3

Vide (4),

The assumptions are: np ≥ 10 and np(1 - p) ≥ 10. Answer 4

[In the given scenario, n = 400 and p = 0.35 and so np = 140 > 10 and np(1 - p) = 91 > 10]

Part (c)

Let X = number of people out of the 400 who replied ‘yes’ to the question. Then,

X ~ B(400, 0.35) ……………………………………………………………...........................…………….. (7)

So, probability that at least 150 out of the 400 replied ‘yes’

= P(X ≥ 150)

= P[Z ≥ {(150 - 140)/√91}]

     [vide (5) and noting that vide Answers (3) and (4) Normal approximation is possible]

= P(Z ≥ 1.0482)

= 0.1473 [vide (5b)] Answer 5

Part (d)

‘If a different similar random sample is taken, it will have a sample proportion less than 0.32’

is equivalent to saying X < 400 x 0.32 = 128.

So, as in Part (c), the required probability

= P(X < 128)

= P[Z < {(128 - 140)/√91}]

= P(Z < - 1.2579)

= 0.1042 [vide (5b)]

Answer 6

DONE