TI-83/84 PLUS LinRegTTest y=a+bx B#0_and P#0 tb=.5789830508 S=, 1117312583 r2=,5858430396 r=.7654038409

Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter7: Distance And Approximation
Section7.3: Least Squares Approximation
Problem 31EQ
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Number 8
For each of the following problems, if no significance level is given, use 0.05.
8. Cereal Killers The amounts of sugar (grams of sugar per gram of cereal) and calories (per
gram of cereal) were recorded for a sample of 16 different cereals. TI-83/84 Plus calculator
results are shown here. Is there sufficient evidence to support the claim that there is a linear
correlation between sugar and calories in a gram of cereal? Explain.
TI-83/84 PLUS
LinRegTTest
=a+bx
B#0_and p#0
tb=.5789830508
s=, 1117312583
r2=,5858430396
r=.7654038409
Transcribed Image Text:For each of the following problems, if no significance level is given, use 0.05. 8. Cereal Killers The amounts of sugar (grams of sugar per gram of cereal) and calories (per gram of cereal) were recorded for a sample of 16 different cereals. TI-83/84 Plus calculator results are shown here. Is there sufficient evidence to support the claim that there is a linear correlation between sugar and calories in a gram of cereal? Explain. TI-83/84 PLUS LinRegTTest =a+bx B#0_and p#0 tb=.5789830508 s=, 1117312583 r2=,5858430396 r=.7654038409
Expert Solution
Step 1

As given,

The correlation coefficient (r) between the amounts of sugar and calories in a gram of cereal. is 0.7654

r = 0.7654

The sample size is n = 16

 

Now we want to test the significance of the correlation coefficient at a 5% level of significance.

Step 2

The most appropriate test is the significance of the correlation coefficient test.

Let be the population correlation coefficient between the variables (the amounts of sugar and calories in a gram of cereal.).

To test the significance the null hypothesis is given as

H0:ρ=0

That is there no is a significant correlation between the variables.

Vs the alternative hypothesis is given as

H1:ρ0

That is there is a significant correlation between the variables.

 

Now obtaining the test statistic.

The test statistic is given as

t=r1-r2n-2

=0.76541-0.7654216-2

t =  4.45

 

The degree of freedom is given as

Df = n -2

= 14

 

Now obtaining the p-value.

Since the hypothesis is two-tailed the p-value for test statistic t =  4.45 with df = 14  is given as

P-value = 0.000549                                                          Using the MS-EXCEL command (T.DIST.2T( 4.45,14))

 

Decision rule: -

At 5% level of significance

Reject the null hypothesis if p-value < 0.05

Accept otherwise.

Since p-value = 0.000549  < 0.05

We reject the null hypothesis.

Conclusion: -

At 5% level of significance, there is sufficient evidence to conclude that there is a significant correlation between the variables (the amounts of sugar and calories).

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