Question

Skunk Hollow, Arkansas, is proud of their energy conservation program. They want to know how much energy they have saved. A sample of 40 houses reveals an average saving of 47.5 kilowatt hours per week with a standard deviation of 55. Put a 90% confidence limit around your best estimate of the average savings. Could this sample have been drawn from a population with a mean of 35? Upper Limit = Lower Limit = The probability that this sample could be drawn from a population with a mean of 35 . . . . p = .

Answer 1

The 90% confidence limit around the average savings is 33.125 kWh to 61.875 kWh. The sample could not have been drawn from a population with a mean of 35 kWh.

Step-by-step explanation:

To calculate the 90% confidence limit around the average savings, we will use the formula:

Confidence Interval = Average Savings ± (Z * (Standard Deviation / Square Root of Sample Size))

Given that the average savings is 47.5 kWh, the standard deviation is 55, and the sample size is 40, we need to find the value of Z for 90% confidence. From the Z-table, the Z value for 90% confidence is approximately 1.645.

Now, we can calculate the confidence interval:

Confidence Interval = 47.5 ± (1.645 * (55 / √40))

Confidence Interval = 47.5 ± 14.375

Upper Limit = 47.5 + 14.375 = 61.875

Lower Limit = 47.5 - 14.375 = 33.125

The probability that this sample could be drawn from a population with a mean of 35 can be calculated using the t-distribution. However, since the sample size is large (n > 30), we can use the normal distribution. We will calculate the z-score:

Z = (Sample Mean - Population Mean) / (Standard Deviation / Square Root of Sample Size)

Z = (47.5 - 35) / (55 / √40) ≈ 3.032

Since the calculated z-score is greater than the critical value of 1.645 for 90% confidence, we can conclude that this sample could not have been drawn from a population with a mean of 35.