Answer:
a) P(x ≤ 0.44) = 0.02275
b) The probability of obtaining a sample proportion less than or equal to 0.44 is very low (2.275%), hence, it would be unusual to obtain a sample proportion less than or equal to 0.44.
c) P(x ≤ 0.50) = 0.30854
A probability of 30.854% doesn't scream unusual, but it is still not a very high probability. So, it is still slightly unusual to obtain a sample proportion of less than half of the voters that don't support the politician.
Step-by-step explanation:
Given,
p = population proportion that support the politician = 0.52
n = sample size = 200
(np = 104) and [np(1-p) = 49.92] are both greater than 10, So, we can treat this problem like a normal distribution problem.
This is a normal distribution problem with
Mean = μ = 0.52
Standard deviation of the sample proportion in the distribution of sample means = σ = √[p(1-p)/n]
σ = √[0.52×0.48)/200]
σ = 0.035 ≈ 0.04
a) Probability of obtaining a sample proportion that is less than or equal to 0.44. P(x ≤ 0.44)
We first normalize/standardize/obtain z-scores for a sample proportion of 0.44
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ = (0.44 - 0.52)/0.04 = -2.00
To determine the probability of obtaining a sample proportion that is less than or equal to 0.44.
P(x ≤ 0.44) = P(z ≤ -2)
We'll use data from the normal probability table for these probabilities
P(x ≤ 0.44) = P(z ≤ -2) = 0.02275
b) Would it be unusual to obtain a sample proportion less than or equal to 0.44 if the politician's claim is true?
The probability of obtaining a sample proportion less than or equal to 0.44 is 0.02275; that is, 2.275%.
The probability of this occurring is very low, hence, it would be unusual to obtain a sample proportion less than or equal to 0.44.
c) If the claim is true, would it be unusual for less than half of the voters in the sample to support the politician?
Sample proportion that matches half of the voters = 0.50
P(x < 0.50)
We follow the same pattern as in (a)
We first normalize/standardize/obtain z-scores for a sample proportion of 0.50
z = (x - μ)/σ = (0.50 - 0.52)/0.04 = -0.50
To determine the probability of obtaining a sample proportion that is less than 0.50
P(x < 0.50) = P(z < -0.50)
We'll use data from the normal probability table for these probabilities
P(x < 0.50) = P(z < -0.50) = 1 - P(z ≥ -0.50) = 1 - P(z ≤ 0.50) = 1 - 0.69146 = 0.30854
Probability of obtaining a sample proportion of less than half of the voters that support the politician = 0.30854 = 30.854%
This value is still not very high, it would still he unusual to obtain such a sample proportion that don't support the politician, but it isn't as unusual as that calculated in (a) and (b) above.
Hope this Helps!!!
Hope now it's obvious. Regards!