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Single Mean and Confidence Interval using Statistics Assignment Solution


Sample Mean for Computer Insurance

Question 1: Single Mean

An information technology company is required to estimate the average value of its computers for insurance purposes. There are 1,350 employees and each one has been issued a computer. The company requires a 95% confidence interval for the total value of computers in the company.

A random sample of 100 computers gives a sample mean value of £680, with a sample standard deviation of £55.

The insurance company charges a flat-rate premium for a total insurance amount up to £900,000. If the total insurance amount is higher than that, the premium is 10% more. An alternative insurance company charges a flat-rate premium for a total insurance amount up to £1,000,000. If the total insurance amount is higher than that, the premium is 12% more. The flat-rate premiums of both insurance companies are the same.

Answer the following questions:

1. Navigate to the “Confidence interval” tab and calculate the confidence interval for the average price per computer. What do these values mean?

2. Convert the calculated confidence interval to the total value of all computers. What do these values mean?

3. Elaborate on whether the information technology company should remain with its current insurer or switch to the alternative, based on the estimated total value of all the computers.

4. Navigate to the “Sample size” tab and calculate how many more computers need to be included in the sample size to estimate the true mean value per computer to within £10.

Answer:

1. The 95% confidence interval for the average price per computer is [669.22, 690.78]. This interval means that 95% of the time, the interval is given by this will contain the population means of computer prices for 1350 employees if we take the 100-sample large number of times.

2. The confidence interval for the total value of the computer is [903,447.3, 932552.7]. This value represents that there is a 95% chance that the total value of the 1350 computers will be in this range.

3. The insurance company should change the insurance company to the alternate one. This is because the current company will charge 10% more for the amount more than the alternate company. Now, as we see from our computation that we are 95% sure to end up higher than £900,000 but less than £1,000,000. So, it makes sense to pay lower by changing the alternate company.

4. From the calculations, we find out that we need a sample size of 117 for the tolerance to become 10. So, we will need 17 extra computers in the sample. 17 more computers in the sample will ensure that our 95% confidence interval for the sample mean is of length 20 and we are estimating the mean within £10.

Question 2: Single Proportion

A large corporation’s training and development manager wants to determine a 98% confidence interval for the proportion of employees who have enrolled with universities while being employed. A random sample of 990 employees indicates that 198 of them have started programs with universities.

Due to a recent change in its staff training and development policy, the corporation is required to show that at least 25% of its employees have enrolled with universities while being employed.

Answer the following questions:

1. What proportion of the sampled employees have enrolled at a university while being employed?

2. Navigate to the “Confidence interval” tab and, using the calculated proportion, determine the endpoints of the confidence interval as a proportion of employees. What do these values mean?

3. Navigate to the “Sample size” tab and calculate how many more employees should be sampled to estimate the true proportion to within 0.02 (or 2%).

4. Based on your calculations of the confidence interval, has the corporation met the requirements of the recent change in its staff training and development policy?

Answer:

 1. The sample proportion of the employees is 198/990=0.2

 2. The 98% confidence interval for the employees who have enrolled with universities while being employed is [0.17, 0.23]. This interval indicates that there is a 98% chance that the actual number of employees who have enrolled with universities while being employed is between 0.17 and 0.23.

3. The number of employees required for the estimate to be within 0.02 is 2165. Hence, we would require 1175 more employees to make the estimate of proportion within 2%.

4. Based on my calculations, the corporation has not met the requirements of the recent change in its staff training and development policy as the confidence interval does not contain 0.25 which is the requirement.