Imagine a researcher asks a sample of five people to drive two types of cars and rate each of them on a scale of 1 to 25. The researcher wants to know: Is there is a difference in the ratings of the two types of cars. Bellow are the data collected: Type A: 20, 10, 11, 23, 7 Type B: 10, 11, 4, 12, 1 Based on available info the significance level is a= 0.5 and the degree of freedom = 4 3. What do you conclude? Present your answer in APA ------------------- NB* I have attached my work solving up to this point. Please help me answer Based on your conclusion identify and explain the type of error you could be making.
Imagine a researcher asks a sample of five people to drive two types of cars and rate each of them on a scale of 1 to 25. The researcher wants to know: Is there is a difference in the ratings of the two types of cars. Bellow are the data collected: Type A: 20, 10, 11, 23, 7 Type B: 10, 11, 4, 12, 1 Based on available info the significance level is a= 0.5 and the degree of freedom = 4 3. What do you conclude? Present your answer in APA ------------------- NB* I have attached my work solving up to this point. Please help me answer Based on your conclusion identify and explain the type of error you could be making.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Imagine a researcher asks a sample of five people to drive two types of cars and rate each of them on a scale of 1 to 25. The researcher wants to know: Is there is a difference in the ratings of the two types of cars. Bellow are the data collected:
Type A: 20, 10, 11, 23, 7
Type B: 10, 11, 4, 12, 1
Based on available info the significance level is a= 0.5 and the degree of freedom = 4 3. What do you conclude? Present your answer in APA
-------------------
NB* I have attached my work solving up to this point. Please help me answer
Based on your conclusion identify and explain the type of error you could
be making.

Transcribed Image Text:Date. 13 Nov / 2)
Page No.
2.
Null and Altesnative Hypotheus:
The following null and altennative hypothesia
meed to be tested,
Ho i Mg =0
Hai Hg o
This corresponds to a to0-tailed test,
for Lobich a t-test for too pained
samples be used.
Rejection RegioN;
Based on the information provided, the
significance level iA d = 0.05, and
the degrees of freedom ane df=4.
Hence, it in found that the critical value
for this tuO0-dailed test is to =2:776
gnd
d=0.05
df3D4,
The nejection negion for thi's to-
tailed
test in R= {t:t) 2776?

Transcribed Image Text:Date.
Page No.-
Test Statiatica
The t-statintic is computed as
shown in below formula:
6.6
4:722/5
t= 3.125
DeciniON about the null hypothesia:
Since it in obrenved that ItH=3.12s
>te=2.776, itis then Concluded
that
the null hypothesia is srejected.
The p-value in p= 0.0353
since
concluded that the null Ay pothew'h
iA nejertid.
and
P=0.035360.05,it in
ConcluxioN:
94 ia concluded that
the null hypothesia Ho in nejected.
enough
Thenefone, thene lA en
evidence
to claim that popeulation mean lly
is different than Mz ,
0.05 Aignifican.ce lovel.
at
the
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