Testing a claim about the mean of the differences, given two dependent samples. The chart below shows the actual temperatures and the temperatures that were forecast by a local TV meteorologist. He claims that his forecasts are the most accurate in town. Complete the chart (first column is done for you). Then test the claim that the mean difference between the actual temperature and the forecast temperature is zero. Use a 0.05 significance level. Actual 1 -5 -5 23 9 Temperature Forecast 16 16 20 22 15 Temperature Differences -15 1. Enter differences in the third row as data values into your TI30XIIS calculator. Find a and $. (You will need this information to calculate your test statistic.) Complete the following: 3. Claim in symbols alternative hypotheses 4. Null and H, : H, : 5. Type of tail Critical t-value(s):_ 6. 8. Decision Curve 9. Test statistic: t= 10. Provided your work is valid, this leads to the conclusion that we CANNOT reject the null hypothesis. Most likely, by observing the original data values, this is not the outcome you anticipated. Yet, this is a valid conclusion. Can vou explain this outcome?
Testing a claim about the mean of the differences, given two dependent samples. The chart below shows the actual temperatures and the temperatures that were forecast by a local TV meteorologist. He claims that his forecasts are the most accurate in town. Complete the chart (first column is done for you). Then test the claim that the mean difference between the actual temperature and the forecast temperature is zero. Use a 0.05 significance level. Actual 1 -5 -5 23 9 Temperature Forecast 16 16 20 22 15 Temperature Differences -15 1. Enter differences in the third row as data values into your TI30XIIS calculator. Find a and $. (You will need this information to calculate your test statistic.) Complete the following: 3. Claim in symbols alternative hypotheses 4. Null and H, : H, : 5. Type of tail Critical t-value(s):_ 6. 8. Decision Curve 9. Test statistic: t= 10. Provided your work is valid, this leads to the conclusion that we CANNOT reject the null hypothesis. Most likely, by observing the original data values, this is not the outcome you anticipated. Yet, this is a valid conclusion. Can vou explain this outcome?
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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![Testing a claim about the mean of the differences, given
two dependent samples.
The chart below shows the actual temperatures and
the temperatures that were forecast by a local TV
meteorologist. He claims that his forecasts are the
most accurate in town. Complete the chart (first
column is done for you). Then test the claim that the
mean difference between the actual temperature and
the forecast temperature is zero. Use a 0.05
significance level.
Actual
1 -5 -5 23 9
Temperature
Forecast
16 16 20 22 15
Тетрerature
Differences -15
1. Enter differences in the third row as data values
into your TI30XIIS calculator. Find a and %.
(You will need this information to calculate your test
statistic.)
Complete the following:
3. Claim in symbols
alternative hypotheses
4. Null and
H, :
H, :
5. Туре of tail
:
6.
Critical t-value(s):
8. Decision Curve
9. Test statistic: t=
10. Provided your work is valid, this leads to the
conclusion that we CANNOT reject the null hypothesis.
Most likely, by observing the original data values, this is
not the outcome you anticipated. Yet, this is a valid
conclusion. Can you explain this outcome?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F29cd9eeb-c72b-48cd-b80d-b19cf468f6ca%2Fefca0648-ba7e-43ad-b3a6-0a386f6ebc5a%2Fmmanchk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Testing a claim about the mean of the differences, given
two dependent samples.
The chart below shows the actual temperatures and
the temperatures that were forecast by a local TV
meteorologist. He claims that his forecasts are the
most accurate in town. Complete the chart (first
column is done for you). Then test the claim that the
mean difference between the actual temperature and
the forecast temperature is zero. Use a 0.05
significance level.
Actual
1 -5 -5 23 9
Temperature
Forecast
16 16 20 22 15
Тетрerature
Differences -15
1. Enter differences in the third row as data values
into your TI30XIIS calculator. Find a and %.
(You will need this information to calculate your test
statistic.)
Complete the following:
3. Claim in symbols
alternative hypotheses
4. Null and
H, :
H, :
5. Туре of tail
:
6.
Critical t-value(s):
8. Decision Curve
9. Test statistic: t=
10. Provided your work is valid, this leads to the
conclusion that we CANNOT reject the null hypothesis.
Most likely, by observing the original data values, this is
not the outcome you anticipated. Yet, this is a valid
conclusion. Can you explain this outcome?
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