StatCrunch logo (home)

Report Properties
Owner: avidmar23
Created: Jul 24, 2016
Share: yes
Views: 344
Tags:
 
Results in this report
 
Data sets in this report
 
Need help?
To copy selected text, right click to Copy or choose the Copy option under your browser's Edit menu. Text copied in this manner can be pasted directly into most documents with formatting maintained.
To copy selected graphs, right click on the graph to Copy. When pasting into a document, make sure to paste the graph content rather than a link to the graph. For example, to paste in MS Word choose Edit > Paste Special, and select the Device Independent Bitmap option.
You can now also Mail results and reports. The email may contain a simple link to the StatCrunch site or the complete output with data and graphics attached. In addition to being a great way to deliver output to someone else, this is also a great way to save your own hard copy. To try it out, simply click on the Mail link.
Austin Vidmar
Mail   Print   Twitter   Facebook

This is an observational study and it is retrospective. 

In this project I will be studying the top 100 retailers.

The parameter I would like to investigate is the mean of the worldwide sales at these retailers.

I believe that the mean will be greater than 15,000 ($ million)

The parameter I would like to investigate is the proportion of worldwide sales in the USA.

I believe the proportion of the worldwide sales accounted for in the USA will be greater than 60%.

 

Data set 1. Top 100 Retailers 2015   [Info]
To analyze this data, please sign in.

 

I am paying attention to the worldwide retail sales ($ millions) and the USA % of worldwide sales.

 

Mean of the worldwide retail sales ($ millions) = 141,484.67 ($ million) & standard deviation = 187,379.01 ($ million)

USA % of worldwide sales mean = 91.92481 & standard deviation = 14.75048

These numbers show that the top 100 retailers make a very good profit. They average over 140,000 ($ million) even though most are a less than that. The average profit for each retailer is amazing. Also with a mean percent of 91, it shows that all of these retailers are bringing in the most sales. They will probably be in business for a very long time considering 91% of the US sales are coming from these retailers.  

<result1

 

Result 2: Chart group stats 1   [Info]
Right click to copy

A sample of the top 100 retailers in the US are shown. In the sample, the mean of the worldwide sales ($ million) is 141,484.67 ($ million) with a standard deviation of 187,379.01 ($ million). Test the claim that the mean of the sales will be greater than 15,000 ($ million). Using a 0.05 significance level. 

Null hypothesis - Reject the claim that the mean will be equal to 15,000 when the mean is 15,000 (u = 15,000)

Alternative hypothesis - Fail to reject the claim that the mean is 15,000 when it is greater than 15,000 ( u > 15000)

I will be conducting a right tailed test becauase the greater than sign is pointing to the right. 

Sample statistic = 141,484.67

 I would use a normal distribution becuase the sample size is greater than 30. 

I would use a z score because the sample size is greater than 30 and I am using the normal distribution. 

141,484.67-15,000 divided by 187,379.01/square root of 100 

z=6.7502

z-score = .7749

There is not significant evidence to support the claim, the z-score is greater than the significant level. Reject the alternative hypothesis

 

 

A sample of the top 100 retailers in the US are shown. In the sample, the mean of the USA % of worldwide sales is 91.92481 with a standard deviation of 14.75048. Test the claim that the mean of the USA % of sales will be greater than 60%. Using a 0.05 significance level. 

Null hypothesis - Reject the claim that the mean will be equal to 60% when the mean is 60% (u = 60%)

Alternative hypothesis - Fail to reject the claim that the mean is 60% when it is greater than 60% (u > 60%)

I will be conducting a right tailed test becauase the greater than sign is pointing to the right. 

Sample statistic = 91.92481

 I would use a normal distribution becuase the sample size is greater than 30. 

I would use a z score because the sample size is greater than 30 and I am using the normal distribution. 

91.92481-60 divided by 14.75048/squar root of 100

z=21.6432

z-score=1

There is not significant evidence to support the claim, the z-score is greater than the significant level. Reject the alternative hypothesis

 

Result 1: Chart group stats   [Info]
Right click to copy

HTML link:
<A href="https://www.statcrunch.com/5.0/viewreport.php?reportid=60236">Austin Vidmar</A>

Comments
Want to comment? Subscribe
Already a member? Sign in.

Always Learning