**Austin Vidmar**

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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]

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.

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**Result 2: Chart group stats 1**[Info]

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]

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