Insurance Companies track life expectancy information to assist in determing the cost of life insurance policies. Last year the average life expectancy of all policy holders was 77 years. ABI Insurance wants to determine if their clients now have a longer life expectancy, on average, so they randomly sample some of their recently paid policies. The ages of the clients in the sample are shown below.
86 75 83 84 81 77 78 79 79 81
76 85 70 76 79 81 73 74 72 83
a) Based on the sample results, find the 90% cofidence interval and interpret
b) For more accurate cost determination, ABI Insurance wants to estimate the average life expectancy to within one year with 95% cofidence. How many randomly selected recently paid policies would they need to sample?
c) Suppose ABI samples 100 recently paid policies. This sample yields a mean of 77.7 years and a standard deviation of 3.6 years. Find a 90% cofidence interval and interpret.
A confidence interval for a parameter is an interval of numbers within which we expect the true value of the population parameter to be contained. The endpoints of the interval are computed based on sample information. a.With smaller samples (n< 30) distribution called the t distribution must be used. The t distribution is similar to the standard normal distribution but takes a slightly different shape depending on the sample size. In a sense, one could think of the t distribution as a family of distributions for smaller samples. Instead of "Z" values, there are "t" values for confidence intervals which are larger for smaller samples, producing larger margins of error, because small samples are less precise. t values are listed by degrees of freedom (df). Just as with large samples, the t distribution assumes that the outcome of interest is approximately normally distributed. For n < 30 Use the t table with df=n-1 Based on the above sample the mean calculated is 78.6 and Standard Deviation is : 4.48 Putting values in formula we get 90% CI =78.6+-1.729*(4.48/sqrt(20)) 76.87,80.33 b.95% CI = Mean +- t(distribution)*Standard Deviation/sqrt(n) =Based on the above sample the mean calculated is 78.6 and Standard Deviation is : 4.48 =78.6+-2.093*(4.48/sqrt(20)) =(76.50,80.70) Note value of t is calculated from t Table taking df as n-1 |
c. If sample size is greater than 30 below formula is used to 90% cofidence interval :
90% CI = Mean+- 1.645*(Standard Deviation/sqrt(N)
=77.7+-1.645*(3.6/sqrt(100)
=(77.11, 78.29)
It means that with 90% confident that sample mean is between 77.11 and 78.29
Insurance Companies track life expectancy information to assist in determing the cost of life insurance policies....
Insurance companies track life expectancy information to assist in determining the cost of life insurance policies. The insurance company knows that, last year, the life expectancy of its policy holders was 77 years. They want to know if their clients this year have a longer life expectancy, on average, so the company randomly samples some of the recently paid policies to see if the mean life expectancy of policy holders has increased. The insurance company will only change their premium...
Show all work: Insurance companies track life expectancy information to assist in determining the cost of life insurance policies. Last year the average life expectancy of all policyholders was 77 years. ABI Insurance wants to determine if their clients now have a longer life expectancy, on average, so they randomly sample some of their recently paid policies. Suppose ABI samples 100 recently paid policies. This sample yields a mean of 77.7 years and a standard deviation of 3.6 years. Find...
6. Insurance companies track life expectancy information to assist in determining the cost of life insurance policies. They want to know if their clients this year have a longer life expectancy, on average, so the company randomly sampled 20 policyholders to see if the mean life expectancy of policyholders has increased. The mean life expectancy for the random sample was 78.6 years with a sample standard deviation of 5.12 years. The insurance company will only change their premium structure if...
Insurance companies track life expectancy information to assist in determining the cost of life insurance policies. Life expectancy is a statistical measure of average time a person is expected to live, based on a number of demographic factors. Mathematically, life expectancy is the mean number of years of life remaining at a given age, assuming age-specific mortality rates remain at their most recently measured levels. Last year the average life expectancy of all the Life Insurance policyholders in Ontario...
6. A bank with a branch located in Crown Point wants to develop an improved process for serving customers during the noon-to-lpm lunch period. Management decides to first study the waiting time of the current process. Data is collected from a random sample of 15 customer and their waiting time is recorded (in minutes) and is located in the Excel data file for this assignment under the tab CPBank. For comparative purposes, similar data is collected from a bank in...
. Suppose a large insurance company wants to estimate the difference between the average amount of term life insurance purchased per family and the average amount of whole life insurance purchased per family. To obtain an estimate, one of the company's actuaries randomly selects 27 families who have term life insurance only and 29 families who have whole life policies only. Each sample is taken from families in which the leading provider is younger than 45 years of age. Use...
Here is a data set (n = 117) that has been sorted. 56.8 69.8 71.2 73.7 75.5 77. 4 78.7 80.3 81. 8 84. 5 87 88.6 92.4 59.9 70.4 72.2 74 75.6 77.5 78.7 80.5 8 2 85 87.1 88.6 92.7 61.2 70.4 72.3 74.1 75.9 77.7 78.7 80.8 82.2 85.3 87.6 88.9 92.8 62.2 68.4 70.5 70.6 72.4 72.5 74.2 74.3 76.1 76.2 77.8 77. 8 78.9 79.1 | 81 | 81.1 82.2 82.2 86.1 86.5 87.8 87.9...
Can someone please check my work? If I made an error can you please correct me and show me the steps of how to get there? Date 1. A national hal survey asked 207 U.s. 3rd graders what their favorite fruit is. ask a. Identify the variable, favorite pre b. Is the e. What is the implied population? aoz e variablequalitativor quantitative (circle one) 2. Categorize nominal, ordinal, interval, or ratio. these measurements associated with smartphones according to level: a....
M. - N O P Q R S T U a 4 Credit Debit 28,250 28,250 AB E F HT Solution: 5 Part A: 6 Jelly World, Inc., is a purveyor of artisan jellies and is known for its high quality Journal Entries: 7 merchandise. The business started up in January 2015 and has been quite 1 Cash 8 successful. Complete the Journal Entries for all 2019 transaction and adjustment Accounts Receivable 9 information listed below. You do not have...
Using the information from P 2–8, prepare and complete a worksheet similar to Illustration 2A–1. Use the information in the worksheet to prepare an income statement and a statement of shareholders’ equity for 2021 and a balance sheet as of December 31, 2021. Cash dividends paid to shareholders during the year amounted to $6,000. Also prepare the necessary closing entries assuming that adjusting entries have been correctly posted to the accounts. P 2-8 Adjusting entries LO2-6 Excalibur Corpoation ells video...