A survey of 90 families showed that 40 owned at least one gun. Find the 95%

confidence interval of the true proportion of families who own at least one gun.
Example

89.98 to 90.02

To find the confidence interval for the true proportion of families who own at least one gun, we can use the formula:

CI = p̂ ± Z * √( p̂ * (1-p̂) / n)

Where:
CI = Confidence Interval
p̂ = Sample Proportion
Z = Z-Score (corresponding to the desired confidence level)
n = Sample Size

Given:
Sample Proportion (p̂) = 40/90 = 0.4444 (rounded to 4 decimal places)
Sample Size (n) = 90
Desired Confidence Level = 95%

First, we need to find the Z-Score corresponding to the desired confidence level. For a 95% confidence level, the Z-Score is 1.96 (approximately) based on the standard normal distribution table.

Now, let's calculate the confidence interval:

CI = 0.4444 ± 1.96 * √( 0.4444 * (1-0.4444) / 90)

CI = 0.4444 ± 1.96 * √( 0.2475 / 90)

CI = 0.4444 ± 1.96 * √0.00275

CI = 0.4444 ± 1.96 * 0.052424789

CI = 0.4444 ± 0.1025699

Therefore, the 95% confidence interval for the true proportion of families who own at least one gun is approximately:

CI = (0.3418, 0.5470)

To find the 95% confidence interval of the true proportion of families who own at least one gun, we can use the formula for the confidence interval for proportions.

The formula is given by:
Confidence Interval = Sample Proportion ± Margin of Error

To calculate the sample proportion, we divide the number of families who own at least one gun by the total number of families in the survey:
Sample Proportion = Number of Families with at least one gun / Total number of families in the survey

In this case, the survey showed that out of 90 families, 40 owned at least one gun. So the sample proportion is:
Sample Proportion = 40 / 90

To calculate the margin of error, we need to use the formula:
Margin of Error = Critical Value * Standard Error

The critical value can be found using a normal distribution table or a calculator. For a 95% confidence level, the critical value is approximately 1.96.

The standard error is given by:
Standard Error = sqrt( (Sample Proportion * (1 - Sample Proportion)) / Sample Size )

In this case, the sample size is 90 families.

Now, let's calculate the confidence interval.

Step 1: Calculate the sample proportion.
Sample Proportion = 40 / 90 ≈ 0.4444

Step 2: Calculate the standard error.
Standard Error = sqrt( (0.4444 * (1 - 0.4444)) / 90 ) ≈ 0.0513

Step 3: Calculate the margin of error.
Margin of Error = 1.96 * 0.0513 ≈ 0.1006

Step 4: Calculate the confidence interval.
Confidence Interval = Sample Proportion ± Margin of Error
Confidence Interval = 0.4444 ± 0.1006

Therefore, the 95% confidence interval for the true proportion of families who own at least one gun is approximately 0.3438 to 0.5449.

Example of a proportional confidence interval formula:

CI95 = p + or - (1.96)[√(pq/n)]
...where p = x/n, q = 1 - p, and n = sample size.
Note: + or - 1.96 represents 95% confidence interval.

For p in your problem: 40/90 = 0.44
For q: 1 - p = 1 - 0.44 = 0.56
n = 90

I let you take it from here to calculate the interval.