# In working with a standard normal distribution there are several

August 30, 2017

Question
This week you answer a second series of problems on statistical inference. You will use Excel to solve the statistics problems. It is EXTREMELY important that you prepare a written discussion of each problem to accompany the numerical calculations. You should attach the Excel file in your Assignment folder.

Note forproblem 1:

In working with a standard normal distribution there are several variables and statistics that you will use. You always start with an underlying distribution of a random variable that you assume is a standard normal distribution. In problem 1 of the assignment this distribution is the distribution of the annual return of common stocks from 1936 to 2010. A standard normal distribution is summarized by its mean and standard deviation.

To show you how to tackle this problem, I will use slightly different numbers: Mean and SD of 12.4% and 20.6%, respectively and probability of more than 10%. Assuming a standard normal distribution for these annual returns for common stocks, you would like to know what the probability would be if an actual return for common stocks between 1936 and 2010 was greater than or less than a particular return. You must convert the 10% into a “z statistic” using the mean and standard deviation of the actual distribution (12.4% and 20.6%). After you calculate the “z” value equivalent to the 10% return, you go to a table for a cumulative standard normal distribution. The table provides values that represent the area under the standard normal distribution curve for specific values of “z.” Half of the area of a standard normal distribution is above “z” = 0 and half is below “z” =0. Therefore, the probability of an actual return being above the mean is 50% and the probability of an actual return being below the mean is 50%. The value in the standard normal distribution table at the web site below gives you the probability between 0 and your z-value. Remember, the mean of the standard normal distribution is 0. This is equivalent to the mean for the distribution of common stock returns, which is 12.4%.

Here’s an example. Assume that you want to know the probability that the actual return could be less than 33%. The z-value would be 1 [(33%-12.4%)/20.6%]. You look for the value in the table where z = 1.00, which is .3413. This is the probability that the actual return will be between 12.4% (z=0) and 33% (z=1), which is the shaded area on the picture at the top of the table. You are looking for the area under the curve to the left of z=1 (33%). To calculate this, you can take the value in the table, which is .3413 and add it to .5000 (this is area under the curve to the left of z =0. The result is .8413 or 84.13%. This is the probability that the actual return will be less than 33%.

Here’s another example of using the z-table in connection with problem 1. Let’s assume that you want to find the probability that the actual return is greater than 5%, you must first calculate the z value for 5%. This is (5% – 12.4%)/20.6%, which is equal to -.3592 or -.36. The negative number means that the z value is below the mean (z = 0 is the mean for a standard normal distribution). This makes sense because the expected return (mean) is 12.4%, and 5% is less than this. The z-table that we are using shows probability values between the z value calculated based on the return given in the problem and z=0. This is the shaded area in the picture. The value in the table for z = .36 is .1406. This means that the probability that an actual return will be between 5% (z = -.36) and 12.4% (z = 0) is 14.06% (to change a value from decimal to percent you multiply by 100; to change a percent to a decimal you divide by 100). You want to find the probability that the actual return will be greater than 5%. This the entire area under the standard normal curve to the right of z = -.36. This is the sum of the areas from z = -.36 to z = 0 and the area to the right of z = 0. You add the value given in the table, .1406, and .50, which is the value for the area to the right of z = 0. The answer is .1406 + .50 = .6406 or 64.06%. The probability that the actual return is greater than 5 % is 64.06%

Probability

Problem Set #2

1. Suppose that the mean of the annual return for common stocks from 2000 to 2012 was 14.37%, and the standard deviation of the annual return was 35.14%. Suppose also that during the same 12-year time span, the mean of the annual return for long-term government bonds was 0.6%, and the standard deviation was 2.1%. The distributions of annual returns for both common stocks and long-term government bonds are bell-shaped and approximately symmetric in this scenario. Assume that these distributions are distributed as normal random variables with the means and standard deviations given previously.

a. Find the probability that the return for common stocks will be greater than 16.32%.

b. Find the probability that the return for common stocks will be greater than 5.89%.

c. Find the probability that the return for common stocks will be less than 14.37%.

Hint: There are many ways to attack this problem in the HW. If you would like the normal distribution table so you can draw the pictures (my preferred way of learning) then I suggest you bookmark this site:

http://www.statsoft.com/textbook/sttable.html

Confidence Interval Estimation

2.Compute a 95% confidence interval for the population mean, based on the sample 1.5, 1.54, 1.55, 1.51, 0.09, 0.08, 1.55, 0.07, 0.99, 0.98, 1.12, 1.13, 1.00, 1.56, and 1.53. Change the last number from 1.53 to 50 and recalculate the confidence interval. Using the results, describe the effect of an outlier or extreme value on the confidence interval.

Hypothesis Testing

3.The management of the Ceebler Fairy Corporation is considering relocating the corporate office to a new location outside HisWood Forest. Management is concerned that the commute times of the employees to the new office might be too long. The company decides to survey a sample of employees at other companies in the same office forest to see how long these employees are commuting to the office. A sample of 23 employees indicated that the employees are commuting X (bar) = 33 minutes and s = 1 minute, 45 seconds.

a. Using the 0.01 level of significance, is there evidence that the population mean is above 32 minutes?

b. What is your answer in (a) if X (bar) = 37 minutes and s = 27 minutes?

c. Look at your answers for a and b above and discuss what you can learn from the results about the effect of a large standard deviation.

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4.Peter’s NEW IT Help company is concerned that the mean wait time of their phone customers for a customer service agent is not greater than 15 minutes. It can be assumed that the population variance is 9 minutes 6 seconds based on past experience. A sample of 563 customers is selected and the sample mean is 16 minutes 30 seconds. Using a level of significance of .10, is there evidence that the population mean wait time is greater than 15 minutes? Fully explain your answer.

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