First we need to create two lists L1 and L2 in TI 84-Plus calculator
Command: Click on STAT >>> 1: Edit
Select L1 and then click on CLEAR
Then enter the given values of first column one by one.
Then select L2 using arrow button
Then enter the given values of second column one by one.
Then select L3
and click on LN >>> 2ND >>> 2 >>> ENTER
So we get the values of L3
Now, we need to run Linear regression test:
Command:
STAT >>> TESTS >>> F : LinRegTTest...ENTER
Look the following image:
Select the input like above image and click on ENTER
So we get the following output
Then using down arrow button we get the remaining output as follows:
From the above output the linear correlation coefficients ( r ) of L1 and L2 is as follows:
r = 0.8895
and p-value = 0.0013
Decision rule:
1) If p-value < level of significance (alpha) then we reject null hypothesis
2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.
Here p value = 0.0013 < 0.05 so we used first rule.
That is we reject null hypothesis
Conclusion: At 5% level of significance there are not sufficient evidence to conclude that the correlation between L1 and L2 is linear.
The linear regression line of L2 on L1 is as follows:
y = a + bx
y = -142253.3889 + 71.5833 x
Let's plug x = 1995 in the above model, we get:
y =-142253.3889+(71.5833* 1995 ) = 555.29 which is approximately equal to 555.
b) Similarly do the regression between L1 and L3
so we get the following result:
From the above output the linear correlation coefficients ( r ) of L1 and L3 is as follows:
r = 0.9963
and p-value = 0.00
Decision rule:
1) If p-value < level of significance (alpha) then we reject null hypothesis
2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.
Here p value = 0.0 < 0.05 so we used first rule.
That is we reject null hypothesis
Conclusion: At 5% level of significance there are not sufficient evidence to conclude that the correlation between L1 and L3 is linear.
The linear regression line of L3 on L1 is as follows:
y = a + bx
y = -913.46 + 0.4614 x
Let's plug x = 1995 in the above model, we get:
y = -913.46 + (0.4614*1995) = 7.033
y = 7.033
Taking exponential of 7.033, we get
y = 1133.43 = 1133
The model in part b ) is better because the value of r in part b) is large than the value of r in part a).
can i get some help on this question please, thanks! The values below shows the number...
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