Question

12 percent of all Nova Scotians are known to have abnormal blood pressure requiring some form of medication.

(give answers to 4 places past decimal)

1. What is the sampling distribution for the sample proportion of Nova Scotians requiring medication in a random sample of 400 Nova Scotians?

Normal with mean:  and standard deviation:

Tries 0/99

2. What is the probability that the sample proportion of Nova Scotians requiring medication, in the random sample of part (1) above, is greater than 14 %?

Tries 0/99

3. What is the minimum sample size necessary for the sample proportion of Nova Scotians requiring medication to be approximately normal?

Answer 1

Answer:

Given,

p = 12% = 0.12

sample n = 400

a)

Mean = up = 0.12

Standard deviation = sqrt(pq/n)

substitute values

= sqrt(0.12(1-0.12)/400)

= 0.0162

b)

P(p^ > 0.14) = P(z > (0.14 - 0.12)/0.0162)

= P(z > 1.23)

= 0.1093486 [since from z table]

= 0.1093

c)

Normally distributed

np >= 5

substitute values

n*0.12 >= 5

n = 5/0.12

n >= 41.66

nq >= 5

n(1 - 0.12) >= 5

n*0.88 >= 5

n >= 5.68

So the minimum sample size n = 42

Answer 2

To answer your questions, we need to use the properties of the binomial distribution and the central limit theorem. Let's calculate the values:

Given information:

• Proportion of Nova Scotians with abnormal blood pressure requiring medication: 12% or 0.12

• Sample size: 400

1. Sampling distribution for the sample proportion: The mean (μ) of the sampling distribution is equal to the population proportion, which is 0.12. The standard deviation (σ) of the sampling distribution can be calculated using the formula:

σ = √((p * (1 - p)) / n)

where p is the population proportion and n is the sample size.

Substituting the values: σ = √((0.12 * (1 - 0.12)) / 400) ≈ 0.0111

Therefore, the sampling distribution for the sample proportion of Nova Scotians requiring medication in a random sample of 400 Nova Scotians is approximately normal with a mean of 0.12 and a standard deviation of 0.0111.

2. Probability that the sample proportion is greater than 14%: To calculate this probability, we need to standardize the sample proportion using the z-score formula:

z = (x - μ) / σ

where x is the value we are interested in, μ is the mean of the sampling distribution, and σ is the standard deviation of the sampling distribution.

Substituting the values: z = (0.14 - 0.12) / 0.0111 ≈ 1.8018

Next, we need to find the probability corresponding to a z-score greater than 1.8018. Using a standard normal distribution table or a statistical software, we find that the probability is approximately 0.0359 or 3.59%.

Therefore, the probability that the sample proportion of Nova Scotians requiring medication in the random sample is greater than 14% is approximately 0.0359 or 3.59%.

3. Minimum sample size for the sample proportion to be approximately normal: To ensure that the sampling distribution of the sample proportion is approximately normal, we typically require a minimum sample size of at least 30. However, for proportions close to 0 or 1, a larger sample size is recommended.

In this case, since the population proportion is 0.12, which is not extremely close to 0 or 1, a sample size of 30 or larger would be sufficient for the sample proportion to be approximately normal.

Therefore, the minimum sample size necessary for the sample proportion of Nova Scotians requiring medication to be approximately normal is 30.