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.
Help me understand how to do this
To find the 99% confidence interval for the population mean number of solicitation calls, we can use the formula:
Confidence Interval = sample mean +/- (critical value * standard deviation / sqrt(sample size))
Let's break down the steps:
Step 1: Determine the critical value.
Since we want a 99% confidence interval, this means we have (1 - 0.99) / 2 = 0.005 in each tail of the distribution. By looking up the Z-score corresponding to a 0.005 tail area (0.995 cumulative probability) in the standard normal distribution table or using a statistical software, we find the critical value to be approximately 2.654.
Step 2: Calculate the confidence interval.
The sample mean is 2.4, the sample standard deviation is 0.9, and the sample size is 14. The formula for the confidence interval becomes:
CI = 2.4 +/- (2.654 * 0.9 / sqrt(14))
Step 3: Solve the confidence interval.
CI = 2.4 +/- (2.654 * 0.9 / sqrt(14))
CI = 2.4 +/- (2.654 * 0.9 / 3.74)
CI = 2.4 +/- (2.499 / 3.74)
CI = 2.4 +/- 0.668
The confidence interval is approximately (1.732, 3.068). This means we are 99% confident that the true population mean number of solicitation calls the O'Sullivan family receives each night falls within this interval.
Interpretation in the context of the problem:
We can interpret the 99% confidence interval as saying that if we were to take many random samples of 14 evenings at the O'Sullivan household and calculate the confidence interval for each sample, approximately 99% of these intervals would contain the true population mean number of solicitation calls. In practical terms, this means that on average, the O'Sullivan family is likely to receive between 1.732 and 3.068 solicitation calls each evening.
Assumptions for using this method:
1. The sample is randomly selected.
2. The distribution of the number of solicitation calls in the population is approximately normal.
3. The sample standard deviation is a reliable estimate of the population standard deviation.
4. The sample size is large enough (n > 30) or the population standard deviation is known.