Question

In: Statistics and Probability

Your co-worker, Jim, randomly draws a sample of 16 of your customers and calculates that 95%...

Your co-worker, Jim, randomly draws a sample of 16 of your customers and calculates that 95% of the businesses have sales between $38,000,000 and $68,000,000. He bases his conclusion on a 95% confidence interval using the properties of the Central Limit Theorem. Jim estimated the standard deviation of the population from the standard deviation of the sample, and then used the z-distribution to calculate the confidence interval.

You have an understanding of statistics principles and immediately see an error in his calculations; What is the error and how can you correct it?

Solutions

Expert Solution

(a)

Error = Central Limit Theorem is valid only for large sample. Here Sample Size =n = 16 < 30. Small Sample. So,Central Limit Theorem cannot be applied. Z distribution is not applicable. Only t test is to be used.

(b)

It an be corrected as follows:

Margin of error (E) is given by:

So,

For = 0.05, Z = 1.96

Thus, we get:

So,

For ndf = 16 - 1 = 15 and = 0.05,

From Table, critical values of t = 2.1314

So

Corrected Confidence Interval:

53,000,000 (2.1314 X 7,653,061.2245)

= 53,000,000 16,311,734.6939

= ( 36,688,265.3061 ,69,311,734.6939)

So,

Corrected Confidence Interval:

36,688,265.31 < < 69,311,734.69


Related Solutions

You and a co-worker rarely see eye to eye. Tomorrow, your co-worker will yell at you...
You and a co-worker rarely see eye to eye. Tomorrow, your co-worker will yell at you for the first time. With what tools will you approach this confrontation? How will the tools that you choose impact your response?
A random sample of 16 pharmacy customers showed the waiting times below (in minutes). 26 16...
A random sample of 16 pharmacy customers showed the waiting times below (in minutes). 26 16 25 18 17 24 18 23 14 20 10 18 19 13 17 16 Click here for the Excel Data File Find a 90 percent confidence interval for μ, assuming that the sample is from a normal population. (Round your standard deviation answer to 4 decimal places and t-value to 3 decimal places. Round your answers to 3 decimal places.)    The 90% confidence...
Maria has data from a random sample of 16 subjects and is constructing a 95% confidence...
Maria has data from a random sample of 16 subjects and is constructing a 95% confidence interval for the population mean. Which value should she use for t*, her critical value?
Suppose a sample of 16 light trucks is randomly selected off the assembly line. The trucks...
Suppose a sample of 16 light trucks is randomly selected off the assembly line. The trucks are driven 1000 miles and the fuel mileage (MPG) of each truck is recorded. It is found that the mean MPG is 22 with a SD equal to 3. The previous model of the light truck got 20 MPG. Questions: a) State the null hypothesis for the problem above b) Conduct a test of the null hypothesis at p = .05. BE SURE TO...
In a cereal box filling factory a randomly selected sample of 16 boxes have an average...
In a cereal box filling factory a randomly selected sample of 16 boxes have an average weight of 470 g with a standard deviation of 15g. The weights are normally distributed. Compute the 90% confidence interval for the weight of a randomly selected box. [I think I can compute the confidence interval but I'm being thrown off by the "randomly-selected box" requirement].
A random sample of 16 pharmacy customers showed the waiting times below (in minutes). Also, the...
A random sample of 16 pharmacy customers showed the waiting times below (in minutes). Also, the sample is from a normal population. Note that σ is unknown. 21 22 22 17 21 17 23 20 20 24 9 22 16 21 22 21 The sample mean is 19.875 and the sample standard deviation is 3.65. Which of the following represents the 80 percent confidence interval for µ? Select one: a. [13.75, 25.25] b. [18.65, 21.10] c. [19.55, 20.425] d. [18.8,...
Determine the mean and standard deviation of your sample. Find the 80%, 95%, and 99% confidence...
Determine the mean and standard deviation of your sample. Find the 80%, 95%, and 99% confidence intervals. Make sure to list the margin of error for the 80%, 95%, and 99% confidence interval. Create your own confidence interval (you cannot use 80%, 95%, and 99%) and make sure to show your work. Make sure to list the margin of error. What trend do you see takes place to the confidence interval as the confidence level rises? Explain mathematically why that...
Determine the mean and standard deviation of your sample. Find the 80%, 95%, and 99% confidence...
Determine the mean and standard deviation of your sample. Find the 80%, 95%, and 99% confidence intervals. Make sure to list the margin of error for the 80%, 95%, and 99% confidence interval. Create your own confidence interval (you cannot use 80%, 95%, and 99%) and make sure to show your work. Make sure to list the margin of error. Problem Analysis—Write a half-page reflection. What trend do you see takes place to the confidence interval as the confidence level...
Determine the mean and standard deviation of your sample. Find the 80%, 95%, and 99% confidence...
Determine the mean and standard deviation of your sample. Find the 80%, 95%, and 99% confidence intervals. Make sure to list the margin of error for the 80%, 95%, and 99% confidence interval. Create your own confidence interval (you cannot use 80%, 95%, and 99%) and make sure to show your work. Make sure to list the margin of error. What trend do you see takes place to the confidence interval as the confidence level rises? Explain mathematically why that...
Your co-worker and you agree that the Fama-French five-factor model explains stock returns better than the...
Your co-worker and you agree that the Fama-French five-factor model explains stock returns better than the market model. Your co-worker ran two sets of 100 time-series regressions of the excess returns of the stocks in the S&P100 index on: 1) the five factors that make up the Fama-French five-factor model and 2) the market factor only. Your co-worker argues that if the five-factor model is indeed better, she should find a smaller number of statistically significant alphas in set 1)...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT