Problem 1

XY Store wants to estimate the average income in an area being considered for expansion using a public survey. Results of that survey you can see in the following table:

We assume the normally distributed household income.

1. Find and interpret a 95% confidence interval for monthly household income in the area!
2. Estimate and interpret the minimal monthly household income in the area! Assume a 90% confidence level!
3. Find a 99% confidence interval for the standard deviation of the income in the area! Interpret it!
   Problem 2
   A random sample of 10 shoppers is selected from Europa Shopping Center to determine the average distance customers travel to the mall. The observed sample (in km) is the following: 5, 1, 15, 20, 3, 4, 5, 4, 0, 7.
   We assume the normally distributed distance.
   1. Find a 90% confidence interval for the expected value of the distance! Interpret it!
   2. Estimate and interpret the maximal distance! Assume a 95% confidence level!
   3. Find a 95% confidence interval for the population standard deviation! Interpret it!

Problem 3

The Fresh-Juice Company packages frozen lemonade in cans claimed to have a mean weight of 0.5 kg. Determine an interval estimate for the population mean and population dispersion if an observed sample of 12 cans have the following weights, in kg:

0.494; 0.504; 0.426; 0.487; 0.503; 0.501; 0.514; 0.495; 0.498; 0.507; 0.483; 0.49.

What should be the range of our random sample if we demand the maximum tolerable error less than 0,001? Assume the normal distribution of the weight!

Problem 4

The length of time (in days) between the billing and the receipt of payment was recorded for 15 randomly selected clients of the legal firms for accounts X, Y and Z. Assume the normally distributed time between the billing and the receipt of payment!

X: 5 6 5 7 6 4 4 5 5 5 5 5 5 5 6

Y: 7 10 7 8 6 8 6 8 10 7 6 7 7 6 8

Z: 4 3 5 5 4 3 3 5 5 4 6 3 2 3 4

1. Calculate and interpret 99% confidence intervals for the mean time between billing and receipt of payment for all firm’s accounts!
2. Do the same thing also for the population dispersion and standard deviation!

Problem 5

A bank wants to estimate the mean amount (in SK) of outstanding consumer loans. The observed sample of 10 outstanding loans is the following

30 000, 25 000, 15 000, 10 000, 28 000, 14 000, 22 000, 17 000, 13 000, 19 000.

Assume the normally distributed amount of loans! Find 95% confidence intervals for the population mean and population dispersion of outstanding loans! Interpret the meaning of these intervals!

Problem 6

The personal director of the XY corporation wants to estimate the number of employees within one year of retirement. A random sample of 300 employee records is observed and 10 people are found to be within one year of retirement. Find a 99% confidence interval for the true proportion of employees within one year of retirement in the entire corporation!

Problem 7

You know percentage daily increases in value of stocks in ten randomly selected days at stock market.

10%, 16%, 5%, 10%, 12%, 8%, 4%, 6%, 5%, 4%.

We assume that the percentage daily increase in value is a random variable with a normal distribution.

1. Find a 95% confidence interval for mean of daily increases in value!
2. Find a 90% confidence interval for standard deviation of daily increases in value!

Price: $36.26

Solution: The downloadable solution consists of 20 pages, 1626 words and 1 charts.
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