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INSTRUCTIONS: Provide (2) 150 words response for RESPONSES 1 AND 2 below. Responses may include direct questions. In your first peer response post, look at the hypothesis test results of one of your classmates and explain what a type 1 error would mean in a practical sense. Looking at your classmate’s outcome, is a type 1 error likely or not? What specific values indicated this?
In your second peer response post, using your classmate’s values, run another hypothesis test using this scenario: A town official claims that the average vehicle in their area Does Not sell for 80th percentile of your data set. Conduct a four-step hypothesis test including a sentence at the end justifying the support or lack of support for the claim and why you made that choice. Note: this test will be different than the initial post, starting with the hypothesis scenario. Use alpha = .05 to test your claim.
Attached are the instructions or word and excel docs for both responses to help with .
Step 1: Use T-test to verify the claim. (No standard deviation) You will need to use the descriptive data from week 2.
What is the Null and Alternative Hypothesis?
Ho: µ = 40th percentile
Ha: µ > 40th percentile
“A town official claims that the average vehicle in their area sells for more than the 40th percentile of your data set.”
- To find this calculation we will use the function =PERCENTILE.INC ()
- =PERCENTILE.INC (E3:E12,0.4) I tried to just put the mean down, but I didn’t get the same results.
- My 40th percentile is 46658
So now my Null and alternative hypothesis is:
Ho: µ = 46658
Ha: µ > 46658
Step 2: Calculate the test statistic: ???????? = ????̅− ????/ ???? √n
- From the data we have from excel
- ????= (Hypothesis Mean)
- ????̅ =56813.7 (Mean of sample)
- n= 10 (sample)
- s= 33061.487 (standard Deviation)
Step 3: Calculate the P-Value.
- To calculate the function for P-value you need =T. DIST.RT() but this is considered a right tailed test.
- Next: Degrees of Freedom (DF)=n-1 (10-1=9)
- My P-Value = T. DIST.RT(0.9704,9) = 0.1786
According to the problem the problem given, the alpha α is 0.05, any my P-value is 0.1786. This means that 0.1786>0.05 (P-value is more than alpha α). So according to the data, we fail to reject the null hypothesis, because P-value is more than alpha α. There isn’t enough data to support the claim, that the average vehicle sells more than the 40th percentile.
(P.S. My new numbers are in Blue and Yellow.) PLEASE FEEDBACK IT WELCOME!!!!
For this week’s discussion I am conducting a t-test. I am conducting a t-test because the sample size is less than thirty and the variance is unknown. We were asked to test the hypothesis that the average vehicle in our vehicle spreadsheets would sell for more than the 40th percentile of our data set. For this test, alpha is 0.05. Here is my four-step hypothesis test:
Step 1: The Hypothesis Scenario
H₀: µ=c or H₀: µ=7690.2
Hₐ: µ>c or Hₐ: µ>7690.2
Based on the use of ‘>’ we know we are using a right tailed test.
Step 2: T-Test Statistic
???????? = (????̅− ????)/(????/√n)*
TS = (18588-7690.2)/6985.884873
TS = 1.560002805
*(????/√n) is the equation for Standard Error which can be found in excel using “=’Standard_Deviation’/SQRT(n) where n is the sample size of 10.
Step 3: P Value
Being a right tailed test, we used the Excel function ‘=T.DIST.RT(TS,n-1)”.
P Value =T.DIST.RT(1.560002805,10-1)
P Value = 0.076595373
Step 4: Conclusion
a) Our P Value is greater than alpha. p>α. 0.0766>0.05.
b) Due to our P Value being greater than alpha, we failed to reject the null hypothesis.
c) There is not enough evidence to support the claim that the average vehicle in the spreadsheet would sell for more than the 40th percentile.
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