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Magazine Chapter 14: Q. 14.133 (page 586)
In Exercises \(14.128-14.133\). we repeat the information from Exercises \(14.22-14.27\) Presuming that the assumptions for regression inferences are met, perform the required correlation \(t-\)tests, using either the critical- value approach or the \(P-\)value approach.
Following are the data on total hours studied over \(2\) weeks and test score at the end of the \(2\) weeks from Exercise \(14.27\).
a. At the \(1%\) significance level, do the data provide sufficient evidence to conclude that a negative linear correlation exists between study time and test score for beginning calculus students?
b. Repeat part (a) using a \(5%\) significance level.
Short Answer
Part a. The null hypothesis is not rejected and it can be conclude that a negative linear correlation exists between study time and test score for beginning calculus students.
Part b. The null hypothesis is not rejected and it can be conclude that a negative linear correlation exists between study time and test score for beginning calculus students.
Step by step solution
Part a. Step 1. Given information
The level of significance is \(0.01\) and the data is,
Part a. Step 2. Calculation
The hypothesis are,
\(H_{0}:\rho=0\)
\(H_{a}:\rho<0\)
The table is shown below.
The value of \(r\) is,
\(r=\frac{\sum x_{i}y_{i}-\sum x_{i}\sum \frac{y_{i}}{n}}{\sqrt{\sum x_{i}^{2}-(\sum x_{i}^{2})}\sqrt{\sum y_{i}^{2}-(\sum y_{i})^{2}}}\)
\(=\frac{9591-(117)\left ( \frac{660}{8} \right )}{\sqrt{1869-\frac{(117)^{2}}{8}}\sqrt{54638-\frac{(660)^{2}}{8}}}\)
\(=-0.774899997\)
The value of test statistic is,
\(t=\frac{r}{\sqrt{\frac{1-r^{2}}{n-2}}}\)
\(=\frac{-0.774899997}{\sqrt{\frac{1-(-0.774899997)^{2}}{8-2}}}\)
\(=-3\)
The degree of freedom is,
\(dof=n-2\)
\(=8-2\)
\(=6\)
The curve is shown below.
Since, the value do not lie in the rejection region.
Thus, the null hypothesis is not rejected and it can be conclude that a negative linear correlation exists between study time and test score for beginning calculus students.
Part b. Step 1. Given information
The level of significance is \(0.05\) and the data is,
Part b. Step 2. Calculation
The hypothesis are,
\(H_{0}:\rho=0\)
\(H_{a}:\rho<0\)
The table is shown below.
The value of \(r\) is,
\(r=\frac{\sum x_{i}y_{i}-\sum x_{i}\sum \frac{y_{i}}{n}}{\sqrt{\sum x_{i}^{2}-(\sum x_{i}^{2})}\sqrt{\sum y_{i}^{2}-(\sum y_{i})^{2}}}\)
\(=\frac{9591-(117)\left ( \frac{660}{8} \right )}{\sqrt{1869-\frac{(117)^{2}}{8}}\sqrt{54638-\frac{(660)^{2}}{8}}}\)
\(=-0.774899997\)
The value of test statistic is,
\(t=\frac{r}{\sqrt{\frac{1-r^{2}}{n-2}}}\)
\(=\frac{-0.774899997}{\sqrt{\frac{1-(-0.774899997)^{2}}{8-2}}}\)
\(=-3\)
The degree of freedom is,
\(dof=n-2\)
\(=8-2\)
\(=6\)
The curve is shown below.
Since, the value do not lie in the rejection region.
Thus, the null hypothesis is not rejected and it can be conclude that a negative linear correlation exists between study time and test score for beginning calculus students.
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