Statistic 6

Sean

Hi everyone,

Well, this week I needed more than just the .pdfs and textbook, I have been spending a lot of time reviewing differences in Z & T- tests. Since I spent so much time on YouTube and reading the book over and over again, I have determined to use a T-test based upon a sample size of less than 30.

The assignment was to determine the sell of vehicles for more than 40th percentile of the vehicles from week 1 which I computed in excel as =PERCENTILE.INC(A1:A10,0.4), with A1:10 as my prices, 0.4 is percentage = $41665.40

Based upon the request to determine a “more than” function, I used a RIGHT SIDED TEST

Step 1:

𝐻0: 𝜇 = 41665.40

𝐻𝑎: 𝜇 > 41665.40

Step 2: & Step 3

Mean (x-bar):

44280.2

Standard Error (SE):

6264.048

Standard Deviation (s):

19808.66

Count (n): 10

𝐻0: 𝜇 = 41665.40

𝐻𝑎: 𝜇 > 41665.40

Sample Error

6264.048

TS: =(44280.20-41665.40)/6264.048 =0.41743

P-Value: =T.DIST.RT(D10,D4-1) = (.41743, 10-1) =0.343073

Step 4:

Conclusion

a.0.343 > .05, Therefore our p-value is greater than alpha.

b.We fail to reject the null hypothesis (Ho) because our p-value is more than alpha

c.There is not enough evidence to suggest that the average vehicle from the type of car you chose during week 1 sells for more than $41665.40.

The WK6 tab on my spreadsheet contains all formulas and data, I can only hope it is correct.

I hope everyone has a good week.

Take care,

Sean

Dylan

Hello class,

For this week’s assignment we were tasked to run a hypothesis test using the data from week 1. The scenario that we were presented with was “A town official claims that the average vehicle in their area sells for more than the 40th percentile of your data set.” The pdf in Handy Helpers was useful in completing this assignment.

First, I determined that I was going to use a t-test. I used a t-test because I was comparing a sample mean with the population mean and it is a small sample group. The next step was to determine the value for the 40th percentile of my week 1 data set. I utilized the excel function =PERCENTILE.INC and highlighted my original data set of 10 cars. This value came to $28,080.

I used a Right Sided Test for my hypothesis. I was testing the claim that the average vehicle in their area sells for more than $28,080. Our professor set the alpha to 0.05. This is the value that will be used to test the following hypotheses:

Ho:u= 28,080- null hypothesis

Ha:u< 28,080- alternative hypothesis

Next, I needed to calculate the test statistic. I started with using excel to find the standard error, SE. I used the following formula: = Sample SD/ SQRT(n) My SE was 4961.37. I used the TS formula from the pdf. The test statistic formula was also performed in excel with the formula =(32,069.50-28,080)/4961.47 with a value of 0.8041.

After that I needed to determine the p-value. The excel function =T.DIST.RT(0.8041,10-1) with a value of 0.22. Based on this calculation we determine that the p-value 0.22 is more than the alpha set to 0.05.

Therefore, we fail to reject the null hypothesis because our p-value is more than the alpha. There is not enough evidence to suggest that the average vehicle sells for more than the 40th percentile of my original data set.

This week was difficult. I did my best to interpret the information in the pdfs and from the videos. I recommend using both. Good Luck.

WEEK 6

40th Percentile of Week 1 Sample $ 28,080.00

Sample Mean (x-bar) $ 32,069.50

SD: $ 15,689.24

Sample Size: 10

Standard Error: $ 4,961.37

TS: 0.8041

p-value 0.22

alpha 0.05

h=28,080 fail to reject null hypothesis because p-value is greater than alpha

h<28080