3. A telephone company wants to estimate the porportion of households that would purchase an additional service line if it were made available at a substantially reduced cost. A random sample of 500 households is selected and the results indicate that 135 of the households would purchase the additional service. Construct and explain a 99% confidence interval estimate of the proportion of all households that would purchase the additional service.

For those of us who don't understand, you didn't help.

To construct a 99% confidence interval estimate of the proportion, we can use the formula:

Confidence interval = sample proportion ± Z-value * √[(sample proportion * (1-sample proportion)) / sample size]

Where:
- The sample proportion is the number of households that would purchase the additional service divided by the sample size.
- The Z-value corresponds to the desired level of confidence. At 99% confidence, the Z-value is approximately 2.576 for a large sample size.
- The sample size is the number of households in the random sample.

First, let's calculate the sample proportion:
Sample proportion = 135/500 = 0.27

Next, we need to calculate the standard error of the proportion:
Standard error of the proportion = √[(0.27 * (1-0.27)) / 500] = 0.0226

Now, we can calculate the margin of error:
Margin of error = 2.576 * 0.0226 = 0.0583

Finally, we can construct the confidence interval:
Confidence interval = 0.27 ± 0.0583
Confidence interval = (0.2117, 0.3283)

Explanation: Based on the given data, we can be 99% confident that the proportion of all households that would purchase the additional service lies within the interval of (0.2117, 0.3283). This means that we estimate with 99% confidence that between 21.17% and 32.83% of all households would purchase the additional service if it were made available at a substantially reduced cost.

To construct a 99% confidence interval estimate of the proportion of all households that would purchase the additional service, we can use the formula:

CI = p̂ ± Z * √(p̂(1-p̂)/n)

Where:
- CI is the confidence interval.
- p̂ is the sample proportion of households that would purchase the additional service.
- Z is the z-score corresponding to the desired confidence level.
- n is the sample size (number of households sampled).

First, we calculate the sample proportion:
p̂ = 135/500 = 0.27

Next, we need to find the z-score for a 99% confidence interval. This can be done using a standard normal distribution table or a calculator. The z-score representing a 99% confidence interval is approximately 2.58.

Substituting the values into the formula:
CI = 0.27 ± 2.58 * √(0.27(1-0.27)/500)

Calculating the expression inside the square root:
√(0.27 * 0.73 / 500) = √(0.1971/500) = √0.0003942 = 0.0199

Plug in the values:
CI = 0.27 ± 2.58 * 0.0199

Calculate the lower and upper bounds of the confidence interval:
Lower Bound = 0.27 - (2.58 * 0.0199) = 0.27 - 0.0513 = 0.2187
Upper Bound = 0.27 + (2.58 * 0.0199) = 0.27 + 0.0513 = 0.3213

Finally, we can conclude that there is a 99% certainty that the proportion of all households that would purchase the additional service lies within the confidence interval of 0.2187 to 0.3213

Example of a proportional confidence interval formula:

CI99 = p + or - (2.58)[√(pq/n)]
...where p = x/n, q = 1 - p, and n = sample size.
Note: + or - 2.58 represents 99% confidence interval.

For p in your problem: 135/500 = 0.27
For q: 1 - p = 1 - 0.27 = 0.73
n = 500

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