# MATH 302 University of Arizona Global Campus Week 6 Statistics Questions

A town official claims that the average vehicle in their area sells for more than the40th percentile of your data set. Using the data, you obtained in week 1, as well as the
summary statistics you found for the original data set (excluding the super car
outlier), run a hypothesis test to determine if the claim can be supported. Make sure
you state all the important values. Use the descriptive statistics you found during
Week 2. Week 2’s SD is in file Stat_22 (1). DO NOT USE the new SD you found
during Week 4. Because again, we are using the original 10 sample data set NOT a
new smaller sample size. Use alpha = .05 to test your claim.
(Note: You will want to use the function =PERCENTILE.INC in Excel to find the
40th percentile of your data set.
First determine if you are using a z or t-test and explain why. Then conduct a fourstep hypothesis test including a sentence at the end justifying the support or lack of
support for the claim and why you made that choice.
Review Week 6 Hypothesis Testing PDF at the bottom of the discussion.
This will give you a step by step example on how to calculate and run a hypothesis
test using Excel. I DO NOT recommend doing this by hand. Let Excel do the heavy
lifting for you.
Instructions: Make sure you include your data set.
Look at the hypothesis test results of file 2 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?
Using values from file 3, 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.
There is an error within week 6 hypothesis testing pdf.
For example 2, the claim is that the average vehicle sells for ‘higher’ than \$23,500, so
the alternative hypothesis (the claim) should be Ha: mu > 23,500 (see below). Sorry
for any confusion! See image below
Useful documents
Week 6 Hypothesis Testing .pdf
STAT_22 (1)
File 1 (Part 1 & 2)
File 2 (Part 1& 2)
H0: Null Hypothesis: μ = 44,505
Ha: Alternate Hypothesis: μ > 44,505
40th percentile: 44,505 (see excel spreadsheet for calculations)
I used the t-test because of the small population size and because the
population variance is unknown.
𝑇𝑆 = 𝑥̅− 𝜇 / 𝑠 / √𝑛 (𝑠 / √𝑛 = Standard Error)
TS = (44725 – 44505)/ 1021.9520/√10 (SE = 323.1696)
TS = 0.6792
I found the p-value by using the excel function T.DIST.RT(0.6792, 101). I chose the right sided or right tailed test because the population
mean (44,725) is higher than the mean defined in the null hypothesis
(44,505).
P-value = 0.2570
Alpha = 0.05
0.2570 > 0.05, the p-value is greater than Alpha.
We fail to reject the null hypothesis H0 because our p-value is more
than Alpha. There is not enough evidence to suggest that the average
vehicle from the type of car chosen during week 1 sells for more than
44,505 (the 40th percentile value).
My hypothesis for this problem is:
𝐻0: 𝜇 = 20,996.4
𝐻1: 𝜇 > 20,996.4
We can find the 40th percentile via: =PERCENTILE.INC(E3:E12,0.4) with
the data range being our price range. This gives me \$20,996.4, which
we will use for our hypothesis test above.
I got my standard error and mean from the week 2 descriptive
statistics.
My Standard error is 374.79, the Mean is \$21,723.1
The Test Statistic is calculated with (the mean – 40th percentile value) /
Standard Error
=1.9389
After that we can obtain our p-value with T.DIST.RT(TS or 1.9389,10-1) :
=0.04226
We are using a T- instead of Z-test here becausewe would need the
pop. variance and we know our SD.
Since we are calculating with an alpha of 0.05, our p-value of ~0.04 falls
below that, meaning we will reject the null hypothesis. I’ll attach the
spreadsheet down below, but the data set for the prices is as follows:
21,000
23,994
20,998
20,990
21,590
20,688
23,500
22,487
20,994
20,990
Vehicle
type/class
Qualititative
Truck (Used)
Truck (Used)
Truck (Used)
Truck (Used)
Truck (Used)
Truck (Used)
Truck (Used)
Truck (Used)
Truck (Used)
Truck (Used)
Year
Make
Model
Qualititative
2013
2013
2013
2012
2014
2012
2014
2015
2013
2014
Qualititative
RAM
RAM
FORD
GMC
RAM
FORD
CHEVROLET
GMC
CHEVROLET
RAM
Qualititative
1500 Express
1500 Express
F150 XLT
Canyon SLE
F150 Lariat
Canyon SLE
1500 Express
Without Supercar
Mean
21723,1
Standard Error
374,7910959
Median
20999
Mode
20990
Standard Deviation
1185,19351
Sample Variance
1404683,656
Kurtosis
0,025759385
Skewness
1,228093971
Range
3306
Minimum
20688
Maximum
23994
Sum
217231
Count
10
Largest(1)
23994
Smallest(1)
20688
Confidence Level(95%)
847,8363604
40th percentile:
Standard Error (from
descriptive statistics):
Test Statistic (TS):
P-value:
Since alpha == 0.05, reject null hypthosis because ou
Our Hypothesis’ were:
Price
MPG (city)
Quantitative
21.000
23.994
20.998
20.990
21.590
20.688
23.500
22.487
20.994
20.990
Quantitative
14
14
14
17
14
14
16
18
13
15
MPG
(highway)
Quantitative
19
19
19
23
20
19
22
25
18
21
20996,4
374,7910959
1,938946811
0,042225959
== 0.05, reject null hypthosis because our p-value of 0.0422 is less than alpha of 0.05
𝐻
𝐻
0:
1:
𝜇
𝜇
= 20,996.4
> 20,996.4
Miles
Quantitative
97,720 miles
78,548 miles
110,151 miles
34,969 miles
7,893 miles
132,514 miles
97,925 miles
54,605 miles
112169 miles
141,181 miles
https://www.kbb.com/cars-for-sale/vehicledetails.xhtml?listingId=626140185&zip=53201&referrer=
https://www.kbb.com/cars-for-sale/vehicledetails.xhtml?listingId=622734487&zip=53201&referrer=
https://www.kbb.com/cars-for-sale/vehicledetails.xhtml?listingId=616606289&zip=53201&referrer=
https://www.kbb.com/cars-for-sale/vehicledetails.xhtml?listingId=628124153&zip=53201&referrer=
https://www.kbb.com/cars-for-sale/vehicledetails.xhtml?listingId=628124174&zip=53201&referrer=
https://www.kbb.com/cars-for-sale/vehicledetails.xhtml?listingId=619987246&listingTypes=USED&v
https://www.kbb.com/cars-for-sale/vehicledetails.xhtml?listingId=622044824&zip=53201&referrer=
https://www.kbb.com/cars-for-sale/vehicledetails.xhtml?listingId=622994891&zip=53201&referrer=
https://www.kbb.com/cars-for-sale/vehicledetails.xhtml?listingId=622435041&zip=53201&referrer=
https://www.kbb.com/cars-for-sale/vehicledetails.xhtml?listingId=620896612&zip=53201&referrer=
https://www.kbb.com/
cars-for-
26140185&zip=53201&referrer=%2Fcars-for-sale%2Fsearchresults.xhtml%3Fzip%3D53201%26city%3DMilwaukee%26vehicleStyleCodes%
22734487&zip=53201&referrer=%2Fcars-for-sale%2Fsearchresults.xhtml%3Fzip%3D53201%26city%3DMilwaukee%26vehicleStyleCodes%
16606289&zip=53201&referrer=%2Fcars-for-sale%2Fsearchresults.xhtml%3Fzip%3D53201%26city%3DMilwaukee%26vehicleStyleCodes%
28124153&zip=53201&referrer=%2Fcars-for-sale%2Fsearchresults.xhtml%3Fzip%3D53201%26city%3DMilwaukee%26vehicleStyleCodes%
28124174&zip=53201&referrer=%2Fcars-for-sale%2Fsearchresults.xhtml%3Fzip%3D53201%26city%3DMilwaukee%26vehicleStyleCodes%
22044824&zip=53201&referrer=%2Fcars-for-sale%2Fsearchresults.xhtml%3Fzip%3D53201%26city%3DMilwaukee%26vehicleStyleCodes%
22994891&zip=53201&referrer=%2Fcars-for-sale%2Fsearchresults.xhtml%3Fzip%3D53201%26city%3DMilwaukee%26vehicleStyleCodes%
22435041&zip=53201&referrer=%2Fcars-for-sale%2Fsearchresults.xhtml%3Fzip%3D53201%26city%3DMilwaukee%26vehicleStyleCodes%
20896612&zip=53201&referrer=%2Fcars-for-sale%2Fsearchresults.xhtml%3Fzip%3D53201%26city%3DMilwaukee%26vehicleStyleCodes%
kee%26vehicleStyleCodes%3DTRUCKS%26incremental%3Dall%26listingTypes%3DUSED%26minPrice%3D20000%26sortBy%3Drelevance%
kee%26vehicleStyleCodes%3DTRUCKS%26incremental%3Dall%26listingTypes%3DUSED%26minPrice%3D20000%26sortBy%3Drelevance%
kee%26vehicleStyleCodes%3DTRUCKS%26incremental%3Dall%26listingTypes%3DUSED%26minPrice%3D20000%26sortBy%3Drelevance%
kee%26vehicleStyleCodes%3DTRUCKS%26incremental%3Dall%26listingTypes%3DUSED%26minPrice%3D20000%26sortBy%3Drelevance%
kee%26vehicleStyleCodes%3DTRUCKS%26incremental%3Dall%26listingTypes%3DUSED%26minPrice%3D20000%26sortBy%3Drelevance%
kee%26vehicleStyleCodes%3DTRUCKS%26incremental%3Dall%26listingTypes%3DUSED%26minPrice%3D20000%26sortBy%3Drelevance%
kee%26vehicleStyleCodes%3DTRUCKS%26incremental%3Dall%26listingTypes%3DUSED%26minPrice%3D20000%26sortBy%3Drelevance%
kee%26vehicleStyleCodes%3DTRUCKS%26incremental%3Dall%26listingTypes%3DUSED%26minPrice%3D20000%26sortBy%3Drelevance%
kee%26vehicleStyleCodes%3DTRUCKS%26incremental%3Dall%26listingTypes%3DUSED%26minPrice%3D20000%26sortBy%3Drelevance%
%26sortBy%3Drelevance%26location%3D%255Bobject%2BObject%255D%26maxPrice%3D24000%26state%3DWI%26searchByDma%3Dfa
%26sortBy%3Drelevance%26location%3D%255Bobject%2BObject%255D%26maxPrice%3D24000%26state%3DWI%26searchByDma%3Dfa
%26sortBy%3Drelevance%26location%3D%255Bobject%2BObject%255D%26maxPrice%3D24000%26state%3DWI%26searchByDma%3Dfa
%26sortBy%3Drelevance%26location%3D%255Bobject%2BObject%255D%26maxPrice%3D24000%26state%3DWI%26searchByDma%3Dfa
%26sortBy%3Drelevance%26location%3D%255Bobject%2BObject%255D%26maxPrice%3D24000%26state%3DWI%26searchByDma%3Dfa
se&sortBy=relevance&numRecords=25&dma=&referrer=%2Fcars-for-sale%2Fused%2Ftruck%2Fmilwaukee-wi-53201%2F%3Fdma%3D%2
%26sortBy%3Drelevance%26location%3D%255Bobject%2BObject%255D%26maxPrice%3D24000%26state%3DWI%26searchByDma%3Dfa
%26sortBy%3Drelevance%26location%3D%255Bobject%2BObject%255D%26maxPrice%3D24000%26state%3DWI%26searchByDma%3Dfa
%26sortBy%3Drelevance%26location%3D%255Bobject%2BObject%255D%26maxPrice%3D24000%26state%3DWI%26searchByDma%3Dfa
%26sortBy%3Drelevance%26location%3D%255Bobject%2BObject%255D%26maxPrice%3D24000%26state%3DWI%26searchByDma%3Dfa
%3D20000%26isNewSearch%3Dfalse%26showAccelerateBanner%3Dfalse%26sortBy%3Drelevance%26numRecords%3D25&clickType=list
de1=FORD&modelCode1=F150PICKUP&clickType=listing
H0: Null Hypothesis: μ = 44,505
Ha: Alternate Hypothesis: μ > 44,505
40th percentile: 44,505 (see excel spreadsheet for calculations)
I used the t-test because of the small population size and because the
population variance is unknown.
𝑇𝑆 = 𝑥̅− 𝜇 / 𝑠 / √𝑛 (𝑠 / √𝑛 = Standard Error)
TS = (44725 – 44505)/ 1021.9520/√10 (SE = 323.1696)
TS = 0.6792
I found the p-value by using the excel function T.DIST.RT(0.6792, 101). I chose the right sided or right tailed test because the population
mean (44,725) is higher than the mean defined in the null hypothesis
(44,505).
P-value = 0.2570
Alpha = 0.05
0.2570 > 0.05, the p-value is greater than Alpha.
We fail to reject the null hypothesis H0 because our p-value is more
than Alpha. There is not enough evidence to suggest that the average
vehicle from the type of car chosen during week 1 sells for more than
44,505 (the 40th percentile value).
My hypothesis for this problem is:
𝐻0: 𝜇 = 20,996.4
𝐻1: 𝜇 > 20,996.4
We can find the 40th percentile via: =PERCENTILE.INC(E3:E12,0.4) with
the data range being our price range. This gives me \$20,996.4, which
we will use for our hypothesis test above.
I got my standard error and mean from the week 2 descriptive
statistics.
My Standard error is 374.79, the Mean is \$21,723.1
The Test Statistic is calculated with (the mean – 40th percentile value) /
Standard Error
=1.9389
After that we can obtain our p-value with T.DIST.RT(TS or 1.9389,10-1) :
=0.04226
We are using a T- instead of Z-test here becausewe would need the
pop. variance and we know our SD.
Since we are calculating with an alpha of 0.05, our p-value of ~0.04 falls
below that, meaning we will reject the null hypothesis. I’ll attach the
spreadsheet down below, but the data set for the prices is as follows:
21,000
23,994
20,998
20,990
21,590
20,688
23,500
22,487
20,994
20,990
Vehicle
type/class
Qualititative
Truck (Used)
Truck (Used)
Truck (Used)
Truck (Used)
Truck (Used)
Truck (Used)
Truck (Used)
Truck (Used)
Truck (Used)
Truck (Used)
Year
Make
Model
Qualititative
2013
2013
2013
2012
2014
2012
2014
2015
2013
2014
Qualititative
RAM
RAM
FORD
GMC
RAM
FORD
CHEVROLET
GMC
CHEVROLET
RAM
Qualititative
1500 Express
1500 Express
F150 XLT
Canyon SLE
F150 Lariat
Canyon SLE
1500 Express
Without Supercar
Mean
21723.1
Standard Error
374.7910959
Median
20999
Mode
20990
Standard Deviation
1185.19351
Sample Variance
1404683.656
Kurtosis
0.025759385
Skewness
1.228093971
Range
3306
Minimum
20688
Maximum
23994
Sum
217231
Count
10
Largest(1)
23994
Smallest(1)
20688
Confidence Level(95%)
847.8363604
40th percentile:
Standard Error (from
descriptive statistics):
Test Statistic (TS):
P-value:
Since alpha == 0.05, reject null hypthosis because ou
Our Hypothesis’ were:
Price
MPG (city)
Quantitative
21,000
23,994
20,998
20,990
21,590
20,688
23,500
22,487
20,994
20,990
Quantitative
14
14
14
17
14
14
16
18
13
15
MPG
(highway)
Quantitative
19
19
19
23
20
19
22
25
18
21
20996.4
374.7910959
1.938946811
0.042225959
== 0.05, reject null hypthosis because our p-value of 0.0422 is less than alpha of 0.05
𝐻
𝐻
0:
1:
𝜇
𝜇
= 20,996.4
> 20,996.4
Miles
Quantitative
97,720 miles
78,548 miles
110,151 miles
34,969 miles
7,893 miles
132,514 miles
97,925 miles
54,605 miles
112169 miles
141,181 miles
https://www.kbb.com/cars-for-sale/vehicledetails.xhtml?listingId=626140185&zip=53201&referrer=
https://www.kbb.com/cars-for-sale/vehicledetails.xhtml?listingId=622734487&zip=53201&referrer=
https://www.kbb.com/cars-for-sale/vehicledetails.xhtml?listingId=616606289&zip=53201&referrer=
https://www.kbb.com/cars-for-sale/vehicledetails.xhtml?listingId=628124153&zip=53201&referrer=
https://www.kbb.com/cars-for-sale/vehicledetails.xhtml?listingId=628124174&zip=53201&referrer=
https://www.kbb.com/cars-for-sale/vehicledetails.xhtml?listingId=619987246&listingTypes=USED&v
https://www.kbb.com/cars-for-sale/vehicledetails.xhtml?listingId=622044824&zip=53201&referrer=
https://www.kbb.com/cars-for-sale/vehicledetails.xhtml?listingId=622994891&zip=53201&referrer=
https://www.kbb.com/cars-for-sale/vehicledetails.xhtml?listingId=622435041&zip=53201&referrer=
https://www.kbb.com/cars-for-sale/vehicledetails.xhtml?listingId=620896612&zip=53201&referrer=
https://www.kbb.com/
cars-for-
26140185&zip=53201&referrer=%2Fcars-for-sale%2Fsearchresults.xhtml%3Fzip%3D53201%26city%3DMilwaukee%26vehicleStyleCodes%
22734487&zip=53201&referrer=%2Fcars-for-sale%2Fsearchresults.xhtml%3Fzip%3D53201%26city%3DMilwaukee%26vehicleStyleCodes%
16606289&zip=53201&referrer=%2Fcars-for-sale%2Fsearchresults.xhtml%3Fzip%3D53201%26city%3DMilwaukee%26vehicleStyleCodes%
28124153&zip=53201&referrer=%2Fcars-for-sale%2Fsearchresults.xhtml%3Fzip%3D53201%26city%3DMilwaukee%26vehicleStyleCodes%
28124174&zip=53201&referrer=%2Fcars-for-sale%2Fsearchresults.xhtml%3Fzip%3D53201%26city%3DMilwaukee%26vehicleStyleCodes%
22044824&zip=53201&referrer=%2Fcars-for-sale%2Fsearchresults.xhtml%3Fzip%3D53201%26city%3DMilwaukee%26vehicleStyleCodes%
22994891&zip=53201&referrer=%2Fcars-for-sale%2Fsearchresults.xhtml%3Fzip%3D53201%26city%3DMilwaukee%26vehicleStyleCodes%
22435041&zip=53201&referrer=%2Fcars-for-sale%2Fsearchresults.xhtml%3Fzip%3D53201%26city%3DMilwaukee%26vehicleStyleCodes%
20896612&zip=53201&referrer=%2Fcars-for-sale%2Fsearchresults.xhtml%3Fzip%3D53201%26city%3DMilwaukee%26vehicleStyleCodes%
kee%26vehicleStyleCodes%3DTRUCKS%26incremental%3Dall%26listingTypes%3DUSED%26minPrice%3D20000%26sortBy%3Drelevance%
kee%26vehicleStyleCodes%3DTRUCKS%26incremental%3Dall%26listingTypes%3DUSED%26minPrice%3D20000%26sortBy%3Drelevance%
kee%26vehicleStyleCodes%3DTRUCKS%26incremental%3Dall%26listingTypes%3DUSED%26minPrice%3D20000%26sortBy%3Drelevance%
kee%26vehicleStyleCodes%3DTRUCKS%26incremental%3Dall%26listingTypes%3DUSED%26minPrice%3D20000%26sortBy%3Drelevance%
kee%26vehicleStyleCodes%3DTRUCKS%26incremental%3Dall%26listingTypes%3DUSED%26minPrice%3D20000%26sortBy%3Drelevance%
kee%26vehicleStyleCodes%3DTRUCKS%26incremental%3Dall%26listingTypes%3DUSED%26minPrice%3D20000%26sortBy%3Drelevance%
kee%26vehicleStyleCodes%3DTRUCKS%26incremental%3Dall%26listingTypes%3DUSED%26minPrice%3D20000%26sortBy%3Drelevance%
kee%26vehicleStyleCodes%3DTRUCKS%26incremental%3Dall%26listingTypes%3DUSED%26minPrice%3D20000%26sortBy%3Drelevance%
kee%26vehicleStyleCodes%3DTRUCKS%26incremental%3Dall%26listingTypes%3DUSED%26minPrice%3D20000%26sortBy%3Drelevance%
%26sortBy%3Drelevance%26location%3D%255Bobject%2BObject%255D%26maxPrice%3D24000%26state%3DWI%26searchByDma%3Dfa
%26sortBy%3Drelevance%26location%3D%255Bobject%2BObject%255D%26maxPrice%3D24000%26state%3DWI%26searchByDma%3Dfa
%26sortBy%3Drelevance%26location%3D%255Bobject%2BObject%255D%26maxPrice%3D24000%26state%3DWI%26searchByDma%3Dfa
%26sortBy%3Drelevance%26location%3D%255Bobject%2BObject%255D%26maxPrice%3D24000%26state%3DWI%26searchByDma%3Dfa
%26sortBy%3Drelevance%26location%3D%255Bobject%2BObject%255D%26maxPrice%3D24000%26state%3DWI%26searchByDma%3Dfa
se&sortBy=relevance&numRecords=25&dma=&referrer=%2Fcars-for-sale%2Fused%2Ftruck%2Fmilwaukee-wi-53201%2F%3Fdma%3D%2
%26sortBy%3Drelevance%26location%3D%255Bobject%2BObject%255D%26maxPrice%3D24000%26state%3DWI%26searchByDma%3Dfa
%26sortBy%3Drelevance%26location%3D%255Bobject%2BObject%255D%26maxPrice%3D24000%26state%3DWI%26searchByDma%3Dfa
%26sortBy%3Drelevance%26location%3D%255Bobject%2BObject%255D%26maxPrice%3D24000%26state%3DWI%26searchByDma%3Dfa
%26sortBy%3Drelevance%26location%3D%255Bobject%2BObject%255D%26maxPrice%3D24000%26state%3DWI%26searchByDma%3Dfa
%3D20000%26isNewSearch%3Dfalse%26showAccelerateBanner%3Dfalse%26sortBy%3Drelevance%26numRecords%3D25&clickType=list
de1=FORD&modelCode1=F150PICKUP&clickType=listing
Vehicle type/Class
CAR
CAR
CAR
CAR
CAR
CAR
CAR
CAR
CAR
CAR
Qualitative
Year
2017
2018
2019
2020
2015
2020
2019
2019
2020
2018
Qualitative
Price
Mean
Standard Error
Median
Mode
Standard Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
Confidence Level(95.0%)
43035.7
692.8271237
43442.5
#N/A
2190.911736
4800094.233
-1.512994966
-0.066317441
6247
40000
46247
430357
10
1567.28384
Make
BMW
Volkswagen
Honda
Subaru
Mitsubishi
Ford
Chevrolet
Nissan
Dodge
Audi
Qualitative
Model
M2
Golf R
Civic Type R
WRX STI
Lancer Evo
Mustang GT
Camaro 2SS
370Z Nismo
Charger Scat
TT
Qualitative
Mean:
Median:
SD:
Sample size:
Price
\$40,999
\$40,000
\$41,498
\$40,435
\$45,198
\$44,095
\$43,895
\$42,990
\$45,000
\$46,247
Quantitative
\$43,036
\$43,443
2190.91174
10
MPG (city)
21
21
22
16
17
15
16
17
15
23
Quantitative
MPG(highway)Horsepower
26
365 hp
29
292 hp
28
306 hp
22
310 hp
23
303 hp
24
460 hp
24
455 hp
26
350 hp
24
485 hp
30
220 hp
Quantitative Quantitative
Price
Mean
43035.7
Standard Error692.827124
Median
43442.5
Mode
#N/A
Standard Deviation
2190.91174
Sample Variance
4800094.23
Kurtosis
-1.512995
Skewness
-0.0663174
Range
6247
Minimum
40000
Maximum
46247
Sum
430357
Count
10
Confidence Level(95.0%)
1567.28384
A town official claims that the average vehicle in their area sells for more than the
40th percentile of your data set. Using the data, you obtained in week 1, as well as the
summary statistics you found for the original data set (excluding the super car
outlier), run a hypothesis test to determine if the claim can be supported. Make sure
you state all the important values. Use the descriptive statistics you found during
Week 2. Week 2’s SD is in file Stat_22 (1). DO NOT USE the new SD you found
during Week 4. Because again, we are using the original 10 sample data set NOT a
new smaller sample size. Use alpha = .05 to test your claim.
(Note: You will want to use the function =PERCENTILE.INC in Excel to find the
40th percentile of your data set.
First determine if you are using a z or t-test and explain why. Then conduct a fourstep hypothesis test including a sentence at the end justifying the support or lack of
support for the claim and why you made that choice.
Review Week 6 Hypothesis Testing PDF at the bottom of the discussion.
This will give you a step by step example on how to calculate and run a hypothesis
test using Excel. I DO NOT recommend doing this by hand. Let Excel do the heavy
lifting for you.
Instructions: Make sure you include your data set.
-Look at the hypothesis test results of ( file 1 Part 1& 2) 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?
-Using values from ( file 2 Part 1&2), 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.
There is an error within week 6 hypothesis testing pdf.
For example 2, the claim is that the average vehicle sells for ‘higher’ than \$23,500, so
the alternative hypothesis (the claim) should be Ha: mu > 23,500 (see below). Sorry
for any confusion! See image below
Useful documents
Week 6 Hypothesis Testing .pdf
STAT_22 (1)
File 1 (Part 1 & 2)
File 2 (Part 1& 2)
Hypothesis Testing is a decision-making process called a Test of Significance.
Below are the 4 unique parts to Hypothesis Testing.
1) The Hypothesis Scenario. This includes the Null and Alternative scenarios.
a. Ho: Null Hypothesis
Ha or H1: Alternative Hypothesis
2) T- Test Statistic
𝑥̅ − 𝜇
𝑇𝑆 = 𝑠
⁄ 𝑛

Where 𝑥̅ is the sample mean, 𝑠 is the sample standard deviation, 𝑛 is the sample size,
and 𝜇 is the population mean defined in the null hypothesis of step 1
3) P-value. The p-value tells you if there is evidence to suggest that your null hypothesis is
incorrect. You will either reject your null hypothesis, or you fail to reject the null
hypothesis. *NOTE: we NEVER Accept null hypotheses, rather find that there was not
enough evidence to reject a null hypothesis. To find the p-value we will use =T.DIST(),
T.DIST.RT() or T.DIST.2T depending on the direction of the alternative hypothesis. The
degrees of freedom (DF) is n – 1.
4) Conclusion:
a. State whether your p-value is greater than alpha (𝛼) or less than alpha (𝛼).
*NOTE, the value of alpha (𝛼) is defined prior to conducting a hypothesis test.
The most common choices for the value of alpha (𝛼) are 0.05, 0.01, 0.10 and
0.005. Alpha (𝛼) is the acceptable Type 1 error. Type 1 error occurs when a null
hypothesis is rejected, though in reality the null hypothesis should not have been
rejected because the null hypothesis is true. Type 1 error is also called the “false
positive” rate.
b. Decide to Reject the null hypothesis (Ho) if the p-value is less than alpha (𝛼), or
Fail to reject the null hypothesis (Ho) if the p-value is more than alpha (𝛼)
c. If you reject Ho, state that there is evidence for the alternative hypothesis. If you
fail to reject Ho, state that there is not enough evidence to support the
alternative hypothesis. *NOTE: we NEVER Accept or Reject alternative
hypotheses, rather we say that our sample provide evidence to support the
alternative hypothesis, or our sample does not provide enough evidence to
support the alternative hypothesis. The alternative hypothesis is not directly
being tested and therefore can not directly be accepted or rejected!
There are 3 different directions of the alternative hypothesis
1) A Left sided, or a Left Tailed Test. We use a left sided test to examine if the population
mean is lower than the mean defined in the null hypothesis (Ho). A left sided test uses
the following hypothesis scenario:
𝐻0: 𝜇 = 𝑐
𝐻𝑎 : 𝜇 < 𝑐 Where 𝑐 is the number we are testing whether the population mean, 𝜇, is less than. For example if we wanted to know if the population mean, 𝜇, is less than 15, then 𝑐 would be 15 in both the null and alternative hypothesis. To find the p-value for a left sided test, we use =T.DIST() in excel. 2) A Right sided, or a Right Tailed Test. We use a right sided test to examine if the population mean is higher than the mean defined in the null hypothesis (Ho). A right sided test uses the following hypothesis scenario: 𝐻0: 𝜇 = 𝑐 𝐻𝑎 : 𝜇 > 𝑐
Where 𝑐 is the number we are testing whether the population mean, 𝜇, is more than.
For example if we wanted to know if the population mean, 𝜇, is more than 15, then 𝑐
would be 15 in both the null and alternative hypothesis. To find the p-value for a right
sided test, we use =T.DIST.RT() in excel.
3) A Two sided, or a Two Tailed Test. We use a two sided test to examine if the population
mean is different than the mean defined in the null hypothesis (Ho). A two sided test
uses the following hypothesis scenario:
𝐻0: 𝜇 = 𝑐
𝐻𝑎 : 𝜇 ≠ 𝑐
Where 𝑐 is the number we are testing whether the population mean, 𝜇, is different
than. For example if we wanted to know if the population mean, 𝜇, is not 15, then 𝑐
would be 15 in both the null and alternative hypothesis. To find the p-value for a two
sided test, we use =T.DIST.2T() in excel.
Example 1 – A Left Sided Test
Now let’s continue to look at our car price data from Week 1. We know the sample mean, 𝑥̅ =
25,650, the sample standard deviation, 𝑠 = 3,488 and n = 10. A friend claims that the average
vehicle from the type of car you chose during week 1 sells for less than \$27,000. Let us test
our friend’s hypothesis using alpha = 0.05. We will write out the 4 steps to do this.
1) State the null and alternative hypotheses
𝐻0 : 𝜇 = 27,000
𝐻𝑎 : 𝜇 < 27,000 2) Calculate the test statistic 𝑥̅ − 𝜇 𝑇𝑆 = 𝑠 ⁄ 𝑛 √ You may notice that the denominator (bottom of the fraction) looks familiar. 𝑠⁄ is the √𝑛 standard error, SE. I start solving for my test statistic, TS, by finding the standard error, SE: When I press enter, I find that the standard error, SE, is 1,103. Now I can simplify my equation for the test statistic: 𝑥̅ − 𝜇 25,650 − 27,000 𝑇𝑆 = = = −1.2238 𝑆𝐸 1,103 *Notice that I am plugging in 𝜇 = 27,000 while calculating my test statistic. I am using the population mean, 𝜇, from the null hypothesis, Ho. 3) P-value. To find the p-value for a left sided test we use =T.DIST() in Excel. The degrees of freedom (DF) is n – 1. When I press enter, I find that the p-value is 0.1261 4) Conclusion: a. Recall from the problem statement that we set alpha = 0.05. 0.1261 > 0.05, 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 less than \$27,000.
Example 2 – A Right Sided Test
Next, a friend claims that the average vehicle from the type of car you chose during week 1
sells for higher than \$23,500. Let us test our friend’s hypothesis using alpha = 0.05. We will
write out the 4 steps to do this.
1) State the null and alternative hypotheses
𝐻0 : 𝜇 = 23,500
𝐻𝑎 : 𝜇 < 23,500 2) Calculate the test statistic 𝑥̅ − 𝜇 𝑇𝑆 = 𝑠 ⁄ 𝑛 √ You may notice that the denominator (bottom of the fraction) looks familiar. 𝑠⁄ is the √𝑛 standard error, SE. I start solving for my test statistic, TS, by finding the standard error, SE: When I press enter, I find that the standard error, SE, is 1,103. Now I can simplify my equation for the test statistic: 𝑥̅ − 𝜇 25,650 − 23,500 𝑇𝑆 = = = 1.9490 𝑆𝐸 1,103 *Notice that I am plugging in 𝜇 = 23,500 while calculating my test statistic. I am using the population mean, 𝜇, from the null hypothesis, Ho. 3) P-value. To find the p-value for a right sided test we use =T.DIST.RT() in Excel. The degrees of freedom (DF) is n – 1. When I press enter, I find that the p-value is 0.0416 4) Conclusion: a. Recall from the problem statement that we set alpha = 0.05. 0.0416 < 0.05, therefore our p-value is less than alpha. b. We reject the null hypothesis (Ho) because our p-value is less than alpha c. There is enough evidence to suggest that the average vehicle from the type of car you chose during week 1 sells for more than \$23,500. Example 3 – A Two Sided Test Lastly, a friend claims that the average vehicle from the type of car you chose during week 1 will not sell for \$23,500. Let us test our friend’s hypothesis using alpha = 0.05. We will write out the 4 steps to do this. 1) State the null and alternative hypotheses 𝐻0 : 𝜇 = 23,500 𝐻𝑎 : 𝜇 ≠ 23,500 2) Calculate the test statistic 𝑥̅ − 𝜇 𝑇𝑆 = 𝑠 ⁄ 𝑛 √ You may notice that the denominator (bottom of the fraction) looks familiar. 𝑠⁄ is the √𝑛 standard error, SE. I start solving for my test statistic, TS, by finding the standard error, SE: When I press enter, I find that the standard error, SE, is 1,103. Now I can simplify my equation for the test statistic: 𝑥̅ − 𝜇 25,650 − 23,500 𝑇𝑆 = = = 1.9490 𝑆𝐸 1,103 *Notice that I am plugging in 𝜇 = 23,500 while calculating my test statistic. I am using the population mean, 𝜇, from the null hypothesis, Ho. 3) P-value. To find the p-value for a two sided test we use =T.DIST.2T() in Excel. The degrees of freedom (DF) is n – 1. When I press enter, I find that the p-value is 0.0831 4) Conclusion: a. Recall from the problem statement that we set alpha = 0.05. 0.0831 > 0.05, 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 a price different than \$23,500.
*Important note when using =T.DIST.2T() in Excel
When your test statistics (TS) is negative, and you have a two sided test, you need to remove
the negative sign before using T.DIST.2T(). One way to remove the negative sign is to use the
absolute value function, ABS(). For example, if I wanted to test whether the average car price
is different than 27,000, the test statistics would be -1.2238. The p-value for this test can be
calculated in Excel as follows:
When I press enter, I find that the p-value is 0.2521. If you forget to take the absolute value
Excel will give you #NUM!. It is important to use the absolute value only for the two sided
t-test.

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