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

1500 Tradesman

F150 Lariat

Silverado 1500 LT

Canyon SLE

Silverado 1500 LT

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

Link:

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=

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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%

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%26sortBy%3Drelevance%26location%3D%255Bobject%2BObject%255D%26maxPrice%3D24000%26state%3DWI%26searchByDma%3Dfa

WI%26searchByDma%3Dfalse%26firstRecord%3D0%26marketExtension%3Dinclude%26relevanceConfig%3Ddefault%26searchRadius%3D

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53201%2F%3Fdma%3D%26searchRadius%3D50%26location%3D%26marketExtension%3Dinclude%26minPrice%3D20000%26isNewSearc

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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

1500 Tradesman

F150 Lariat

Silverado 1500 LT

Canyon SLE

Silverado 1500 LT

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

Link:

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/

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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%

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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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