Question

math problem Just before a referendum on a school budget, a local newspaper polls 400 voters in an attempt to predict whether the budget will pass. Suppose that the budget actually has the support of 52% of the voters. What’s the probability the newspaper’s sample will lead them to predict defeat? Be sure to verify that the assumptions and conditions necessary for your analysis are

Answer 1

The conditions necessary for the analysis are met.

The probability the newspaper’s sample will lead them to predict defeat is 0.7881

Explanation:

We are given;

population proportion; μ = 52% = 0.52

Sample size;n = 400

The conditions are;

10% conditon: sample size is less than 10% of the population size

Success or failure condition; np = 400 x 0.52 = 208 and n(1 - p) = 400(1 - 0.52) = 192.

Both values are greater than 10

Randomization condition; we assume that the voters were randomly selected.

So the conditions are met.

Now, the standard deviation is gotten from;

σ = √((p(1 - p)/n)

where;

p is the population proportion

n is the sample size

σ is standard deviation

Thus;

σ = √((0.52(1 - 0.52)/400)

σ = √((0.52(0.48)/400)

σ = 0.025

Now to find the z-value, we'll use;

P(p^ > 0.5) = P(z > (x - μ)/σ)

Thus;

P(p^ > 0.5) = P(z > (0.5 - 0.52)/0.025)

This gives;

P(p^ > 0.5) = P(z > - 0.8)

This gives;

P(p^ > 0.5) = 1 - P(z < -0.8)

From the table attached we have a z value of 0.21186

Thus;

P(p^ > 0.5) = 1 - 0.21186 = 0.7881

Thus, the probability the newspaper’s sample will lead them to predict defeat is 0.7871

Answer 2

The probability, P that the newspaper's sample will lead them to predict defeat is 0.21186

Explanation:

Here we have

p = 52%, therefore np = 208 Voters

q = 1 - p = 1 - 0.52 = 0.48

nq = 0.48×400 = 192 Voters

Therefore, as np and nq are both > 10 the conditions for approximation to normality are met

We therefore have

μ = 0.52 and

σ =
`\sqrt{(pq)/(n) } = \sqrt{(0.52 * 0.48)/(400) } = 0.0249799 \approx 0.025`

The z score is therefore;

`Z=(x-\mu )/(\sigma ) =(0.5-0.52 )/(0.025 )= -0.8`

The probability the newspaper's sample will lead them to predict defeat is given by P(Z < -0.8) = 0.21186.