A random sample of 14 evenings at the O'Sullivan house hold showed that the family received an average of 2.4 solicitation calls each evening. The sample standard deviation was s= 0.9. Find a 99% confidence interval for the population mean number of solicitation calls this family receives each night. Assume that the distribution of the number of solicitation calls this family distribution. Interpret the interval in the context of this problem. State what assumptions were met to use the method you selected.

To find the 99% confidence interval for the population mean number of solicitation calls the O'Sullivan household receives each night, we can use the formula for a confidence interval:

Confidence Interval = sample mean +/- (critical value * standard deviation / sqrt(sample size))

First, let's find the critical value. Since we want a 99% confidence level, we need to find the z-score that corresponds to that confidence level. We can use a z-table or a calculator to find the critical value. In this case, the critical value for a 99% confidence level is 2.896.

Next, we plug in the values into the formula:

Confidence Interval = 2.4 +/- (2.896 * 0.9 / sqrt(14))

The sample mean is 2.4, the standard deviation is 0.9, and the sample size is 14.

Calculating the expression inside the parentheses:

2.896 * 0.9 = 2.6064

Calculating the square root of the sample size:

sqrt(14) = 3.7417

Now we can substitute these values back into the formula:

Confidence Interval = 2.4 +/- (2.6064 / 3.7417)

Simplifying:

Confidence Interval = 2.4 +/- 0.6967

The lower limit of the confidence interval is 2.4 - 0.6967 = 1.7033
The upper limit of the confidence interval is 2.4 + 0.6967 = 3.0967

Therefore, the 99% confidence interval for the population mean number of solicitation calls the O'Sullivan family receives each night is (1.7033, 3.0967).

Interpretation in the context of this problem:
We are 99% confident that the true population mean number of solicitation calls the O'Sullivan family receives each night is between 1.7033 and 3.0967. This means that on average, the family can expect to receive between approximately 1.7 and 3.1 solicitation calls each evening.

Assumptions made for this method:
1. The sample is randomly selected from the population.
2. The distribution of the number of solicitation calls follows a normal distribution.
3. The sample standard deviation (s=0.9) is a good estimate of the population standard deviation (σ).