suppose we wish to construct a confidence interval for a population proportion p. if we sample without replacement from a relatively small population of size N, the margin or error E is modified to include the finite population correction factor as follows:

E=Za/2 pq/n N-n/N-1

construct a 90% confidence interval for the porportion of students at a school who are left handed. the numbers of students at the school is N=320. in a random sample of 80 students, selected without replacement, there are 7 left handers.

To construct a confidence interval for a population proportion, you can use the formula:

E = Z * sqrt((p*q) / n)

Where:
- E is the margin of error
- Z is the z-score corresponding to the desired level of confidence
- p is the sample proportion
- q is 1 - p
- n is the sample size

However, when sampling without replacement from a relatively small population, we need to account for the finite population correction factor. The formula with correction becomes:

E = (Z * sqrt((p*q) / n)) * sqrt((N-n) / (N-1))

Where:
- N is the population size
- n is the sample size

In this case, you want to construct a 90% confidence interval for the proportion of left-handed students at a school with a population size of N = 320. In a random sample of 80 students, there are 7 left-handed students.

To calculate the confidence interval, we first need to find the sample proportion (p) and the standard deviation (σ).

p = number of successes / sample size
p = 7/80 ≈ 0.0875

q = 1 - p
q = 1 - 0.0875 ≈ 0.9125

Next, we find the z-score corresponding to the desired level of confidence. For a 90% confidence interval, the z-score is approximately 1.645.

Finally, we can calculate the margin of error (E) using the formula:

E = (Z * sqrt((p*q) / n)) * sqrt((N-n) / (N-1))
E = (1.645 * sqrt((0.0875 * 0.9125) / 80)) * sqrt((320-80) / 319)
E ≈ 0.058

The confidence interval is then given by the sample proportion plus or minus the margin of error:

Confidence Interval = p ± E
Confidence Interval ≈ 0.0875 ± 0.058

Therefore, the 90% confidence interval for the proportion of left-handed students at the school is approximately (0.0295, 0.1455).

To construct a 90% confidence interval for the proportion of left-handed students at the school, we can follow these steps:

Step 1: Calculate the sample proportion.
Given that in a random sample of 80 students, there are 7 left-handers, we can calculate the sample proportion (p̂) as the number of left-handers divided by the sample size:
p̂ = 7/80 = 0.0875

Step 2: Calculate the finite population correction factor.
The finite population correction factor is given by the formula N-n/N-1, where N is the population size and n is the sample size. In this case, N = 320 (number of students at the school) and n = 80 (sample size), so the finite population correction factor is:
N-n/N-1 = 320-80/320-1 = 240/319 ≈ 0.752

Step 3: Determine the critical value.
For a 90% confidence interval, we need to find the critical value (Za/2) from the standard normal distribution. We can use a probability table, calculator, or statistical software to find this value. The critical value for a 90% confidence interval is approximately 1.645.

Step 4: Calculate the margin of error (E).
The margin of error (E) is calculated using the following formula:
E = Za/2 * √(p̂q/n) * (N-n/N-1)
where q = 1 - p̂ (complement of the sample proportion).

In this case, we have:
E = 1.645 * √(0.0875 * 0.9125/80) * 0.752
E ≈ 0.0912

Step 5: Calculate the confidence interval.
The confidence interval can be calculated as:
Lower bound = p̂ - E
Upper bound = p̂ + E

In this case:
Lower bound = 0.0875 - 0.0912 ≈ -0.0037 (since the lower bound cannot be negative, it should be rounded to 0)
Upper bound = 0.0875 + 0.0912 ≈ 0.1787

The 90% confidence interval for the proportion of left-handed students at the school is approximately 0 to 0.179, or [0, 0.179].