Hello,
I have a few questions I'm working on for Statistics hw. Here's one of them - please help me, I'm stuck.
It is known that 2% of the population has a general injection allergy. Suppose the Health Dept. randomly selected 250 individuals from the population and there were 7 people known to have the allergy .
a. construct a 99% confidence interval for the population proportion p.
I can't figure this out because i don't have the numbers or the standard deviation. i was thinking p(1-p)/n, which gives .0089 when I find the root. then is it P(Z=.15) = P(z=14.61?) The problem is that I can't find this number on my table of z scores.
b. what is the probability that the proportion in the sample that has the injection allergy will be as great as 4%?
Is this the same as above but P(Z=.04)?
Thank you for you help!
Formula:
CI99 = p + or - (2.58)(sqrt of pq/n)
...where sqrt = square root; p = x/n; q = 1 - p; and + or - 2.58 represents the 99% confidence interval using a z-table.
With your sample data:
CI99 = 7/250 + or - (2.58)[sqrt of (7/250)(243/250)/250]
Convert fractions to decimals.
I'll let you take it from there.
Perhaps someone else can help you with part b.
To construct a confidence interval for the population proportion, you can use the following formula:
CI = p̂ ± z * sqrt(p̂q̂/n)
Where:
- CI is the confidence interval
- p̂ is the sample proportion (in this case, 7/250)
- z is the z-score corresponding to the desired confidence level (for a 99% confidence level, z = 2.58)
- q̂ is the complement of the sample proportion (1 - p̂)
- n is the sample size (in this case, 250)
Let's calculate the confidence interval step by step using your sample data:
1. Calculate p̂: p̂ = 7/250 = 0.028
2. Calculate q̂: q̂ = 1 - p̂ = 1 - 0.028 = 0.972
3. Calculate the standard error: sqrt(p̂q̂/n) = sqrt((0.028 * 0.972) / 250) ≈ 0.0122
4. Calculate the margin of error: z * sqrt(p̂q̂/n) = 2.58 * 0.0122 ≈ 0.0315
5. Calculate the lower bound of the confidence interval: p̂ - (margin of error) = 0.028 - 0.0315 ≈ -0.0035 (Note that the lower bound cannot be negative, so we take it as 0)
6. Calculate the upper bound of the confidence interval: p̂ + (margin of error) = 0.028 + 0.0315 ≈ 0.0595
Hence, the 99% confidence interval for the population proportion p is approximately [0, 0.0595].
For part b, you want to find the probability that the proportion in the sample that has the injection allergy will be as great as 4%. This can be calculated using the z-score and the formula for the standard normal distribution:
P(x ≥ 0.04) = 1 - P(x < 0.04) = 1 - P(z < (0.04 - p̂) / sqrt(p̂q̂/n))
Plugging in the values from your question:
P(x ≥ 0.04) = 1 - P(z < (0.04 - 0.028) / sqrt((0.028 * 0.972) / 250))
Now, you can look up the value (0.04 - 0.028) / sqrt((0.028 * 0.972) / 250) in the z-table to find the corresponding probability.