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Chapter 9: Problem 2

A random sample of size \(n=6\) from a bivariate normal distribution yields a value of the correlation coefficient of \(0.89 .\) Would we accept or reject, at the 5 percent significance level, the hypothesis that \(\rho=0\).

Short Answer

Expert verified
The decision to reject or accept the hypothesis that \(\rho=0\) would be based on the comparison of the test statistic (t-score) with the critical t-value. This complete answer would be provided once the values are calculated using the given sample size and correlation coefficient.

Step by step solution

01

Calculate Test Statistic (t-score)

The formula for test statistic in this case is given by: \[t = (r * \sqrt{n-2}) / (\sqrt{1-r^2})\]. Here, 'r' represents the sample correlation coefficient and 'n' is the sample size. Substituting the given values, compute the test statistic.
02

Compute the Degrees of Freedom

The degrees of freedom is given by 'n-2', where 'n' is the sample size. So, in this case, calculate degrees of freedom as (6-2).
03

Find Critical t-value

Now, we need to find the t critical value which will serve as a threshold for our decision. For a 5% significance level with degrees of freedom as calculated in Step 2, look up the t-distribution table or use statistical software to find the critical t-value.
04

Compare Test Statistic and Critical t-value

Now compare the calculated test statistic from Step 1 with the critical t-value from Step 3. If the absolute value of test statistic is greater than the critical t-value, we reject the null hypothesis (\(\rho=0\)), else we fail to reject it.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Correlation Coefficient
Understanding the correlation coefficient is fundamental to grasping bivariate normal distribution testing. It is a numerical measure that describes the strength and direction of a linear relationship between two variables. The coefficient ranges between -1 and 1, where 1 indicates perfect positive correlation, -1 indicates perfect negative correlation, and 0 signifies no linear correlation.

In hypothesis testing, the correlation coefficient is crucial in evaluating the relationship we assume exists between the variables鈥攊n our exercise, whether \(\rho = 0\) or not. A value of 0.89 suggests a strong positive relationship in our sample, leading to further investigation through hypothesis testing to determine if this finding is statistically significant or could have occurred by chance.
Test Statistic (t-score)
The test statistic, particularly the t-score, plays a pivotal role in hypothesis testing. It is essentially the standardized value that helps us assess the strength of the evidence against the null hypothesis. Calculated using the formula: \[t = (r * \sqrt{n-2}) / (\sqrt{1-r^2})\], it incorporates the sample correlation coefficient (r) and the sample size (n) to produce a t-score that we can compare against a known distribution鈥攈ere, the t-distribution.

In essence, the test statistic helps us quantify how far our sample statistic is from the null hypothesis value, measured in relation to the standard error. The calculated t-score from the exercise indicates the statistical significance of the observed correlation.
Critical t-value
The critical t-value serves as a cut-off point or threshold in hypothesis testing. Determined from the t-distribution, it considers both the chosen significance level (such as 0.05 for a 5% significance level) and the degrees of freedom. It essentially sets the boundary for what we would consider an extreme value under the null hypothesis.

If our test statistic exceeds this critical value, this indicates that our sample result is sufficiently extreme to reject the null hypothesis鈥攊mplying the sample statistic is not consistent with the null hypothesis within the chosen level of confidence. Finding the critical t-value requires statistical tables or software, which provide the value tailored to our test's specific requirements.
Degrees of Freedom
Degrees of freedom are an essential concept when performing statistical tests, especially when dealing with any t-distribution based calculations. It is a value that represents the number of independent values or observations that are free to vary when estimating statistical parameters. In our exercise, the degrees of freedom are calculated as the sample size minus two (n - 2).

The reason for using \(n - 2\) in correlation tests is that two parameters are being estimated鈥攂oth the slope and the intercept of the linear relationship. Degrees of freedom are used to describe the shape of the t-distribution, which is necessary for identifying the correct critical t-value. Consequently, as the degrees of freedom increase, the t-distribution approaches the normal distribution. Understanding this helps in interpreting test outcomes and in determining accuracy levels for our statistical conclusions.

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Most popular questions from this chapter

Often in regression the mean of the random variable \(Y\) is a linear function of \(p\) -values \(x_{1}, x_{2}, \ldots, x_{p}\), say \(\beta_{1} x_{1}+\beta_{2} x_{2}+\cdots+\beta_{p} x_{p}\), where \(\boldsymbol{\beta}^{\prime}=\left(\beta_{1}, \beta_{2}, \ldots, \beta_{p}\right)\) are the regression coefficients. Suppose that \(n\) values, \(\boldsymbol{Y}^{\prime}=\left(Y_{1}, Y_{2}, \ldots, Y_{n}\right)\) are observed for the \(x\) -values in \(\boldsymbol{X}=\left[x_{i j}\right]\), where \(\boldsymbol{X}\) is an \(n \times p\) design matrix and its ith row is associated with \(Y_{i}, i=1,2, \ldots, n .\) Assume that \(Y\) is multivariate normal with mean \(\boldsymbol{X} \boldsymbol{\beta}\) and variance-covariance matrix \(\sigma^{2} \boldsymbol{I}\), where \(\boldsymbol{I}\) is the \(n \times n\) identity matrix. (a) Note that \(Y_{1}, Y_{2}, \ldots, Y_{n}\) are independent. Why? (b) Since \(\boldsymbol{Y}\) should approximately equal its mean \(\boldsymbol{X} \boldsymbol{\beta}\), we estimate \(\boldsymbol{\beta}\) by solving the normal equations \(\boldsymbol{X}^{\prime} \boldsymbol{Y}=\boldsymbol{X}^{\prime} \boldsymbol{X} \boldsymbol{\beta}\) for \(\boldsymbol{\beta}\). Assuming that \(\boldsymbol{X}^{\prime} \boldsymbol{X}\) is non- singular, solve the equations to get \(\hat{\boldsymbol{\beta}}=\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)^{-1} \boldsymbol{X}^{\prime} \boldsymbol{Y}\). Show that \(\hat{\boldsymbol{\beta}}\) has a multivariate normal distribution with mean \(\boldsymbol{\beta}\) and variance-covariance matrix $$ \sigma^{2}\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)^{-1} $$ (c) Show that $$ (\boldsymbol{Y}-\boldsymbol{X} \boldsymbol{\beta})^{\prime}(\boldsymbol{Y}-\boldsymbol{X} \boldsymbol{\beta})=(\hat{\boldsymbol{\beta}}-\boldsymbol{\beta})^{\prime}\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)(\hat{\boldsymbol{\beta}}-\boldsymbol{\beta})+(\boldsymbol{Y}-\boldsymbol{X} \hat{\boldsymbol{\beta}})^{\prime}(\boldsymbol{Y}-\boldsymbol{X} \hat{\boldsymbol{\beta}}) $$ say \(Q=Q_{1}+Q_{2}\) for convenience. (d) Show that \(Q_{1} / \sigma^{2}\) is \(\chi^{2}(p)\). (e) Show that \(Q_{1}\) and \(Q_{2}\) are independent. (f) Argue that \(Q_{2} / \sigma^{2}\) is \(\chi^{2}(n-p)\). (g) Find \(c\) so that \(c Q_{1} / Q_{2}\) has an \(F\) -distribution. (h) The fact that a value \(d\) can be found so that \(P\left(c Q_{1} / Q_{2} \leq d\right)=1-\alpha\) could be used to find a \(100(1-\alpha)\) percent confidence ellipsoid for \(\beta\). Explain.

Let \(X_{1}\) and \(X_{2}\) be two independent random variables. Let \(X_{1}\) and \(Y=\) \(X_{1}+X_{2}\) be \(\chi^{2}\left(r_{1}, \theta_{1}\right)\) and \(\chi^{2}(r, \theta)\), respectively. Here \(r_{1}

Let \(\boldsymbol{X}^{\prime}=\left[X_{1}, X_{2}, \ldots, X_{n}\right]\), where \(X_{1}, X_{2}, \ldots, X_{n}\) are observations of a random sample from a distribution which is \(N\left(0, \sigma^{2}\right) .\) Let \(b^{\prime}=\left[b_{1}, b_{2}, \ldots, b_{n}\right]\) be a real nonzero vector, and let \(\boldsymbol{A}\) be a real symmetric matrix of order \(n\). Prove that the linear form \(\boldsymbol{b}^{\prime} \boldsymbol{X}\) and the quadratic form \(\boldsymbol{X}^{\prime} \boldsymbol{A} \boldsymbol{X}\) are independent if and only if \(\boldsymbol{b}^{\prime} \boldsymbol{A}=\mathbf{0}\). Use this fact to prove that \(\boldsymbol{b}^{\prime} \boldsymbol{X}\) and \(\boldsymbol{X}^{\prime} \boldsymbol{A} \boldsymbol{X}\) are independent if and only if the two quadratic forms, \(\left(\boldsymbol{b}^{\prime} \boldsymbol{X}\right)^{2}=\boldsymbol{X}^{\prime} \boldsymbol{b} \boldsymbol{b}^{\prime} \boldsymbol{X}\) and \(\boldsymbol{X}^{\prime} \boldsymbol{A} \boldsymbol{X}\) are independent.

Student's scores on the mathematics portion of the ACT examination, \(x\), and on the final examination in the first-semester calculus ( 200 points possible), \(y\), are given. (a) Calculate the least squares regression line for these data. (b) Plot the points and the least squares regression line on the same graph. (c) Find point estimates for \(\alpha, \beta\), and \(\sigma^{2}\). (d) Find 95 percent confidence intervals for \(\alpha\) and \(\beta\) under the usual assumptions. $$ \begin{array}{cc|cc} \hline \mathrm{x} & \mathrm{y} & \mathrm{x} & \mathrm{y} \\ \hline 25 & 138 & 20 & 100 \\ 20 & 84 & 25 & 143 \\ 26 & 104 & 26 & 141 \\ 26 & 112 & 28 & 161 \\ 28 & 88 & 25 & 124 \\ 28 & 132 & 31 & 118 \\ 29 & 90 & 30 & 168 \\ 32 & 183 & & \\ \hline \end{array} $$

(Bonferroni Multiple Comparison Procedure). In the notation of this section, let \(\left(k_{i 1}, k_{i 2}, \ldots, k_{i b}\right), i=1,2, \ldots, m\), represent a finite number of \(b\) -tuples. The problem is to find simultaneous confidence intervals for \(\sum_{j=1}^{b} k_{i j} \mu_{j}, i=1,2, \ldots, m\), by a method different from that of Scheff茅. Define the random variable \(T_{i}\) by $$ \left(\sum_{j=1}^{b} k_{i j} \bar{X}_{. j}-\sum_{j=1}^{b} k_{i j} \mu_{j}\right) / \sqrt{\left(\sum_{j=1}^{b} k_{i j}^{2}\right) V / a}, \quad i=1,2, \ldots, m $$ (a) Let the event \(A_{i}^{c}\) be given by \(-c_{i} \leq T_{i} \leq c_{i}, i=1,2, \ldots, m\). Find the random variables \(U_{i}\) and \(W_{i}\) such that \(U_{i} \leq \sum_{1}^{b} k_{i j} \mu_{j} \leq W_{j}\) is equivalent to \(A_{i}^{c}\) (b) Select \(c_{i}\) such that \(P\left(A_{i}^{c}\right)=1-\alpha / m ;\) that is, \(P\left(A_{i}\right)=\alpha / m .\) Use Exercise 9.4.1 to determine a lower bound on the probability that simultaneously the random intervals \(\left(U_{1}, W_{1}\right), \ldots,\left(U_{m}, W_{m}\right)\) include \(\sum_{j=1}^{b} k_{1 j} \mu_{j}, \ldots, \sum_{j=1}^{b} k_{m j} \mu_{j}\) respectively. (c) Let \(a=3, b=6\), and \(\alpha=0.05\). Consider the linear functions \(\mu_{1}-\mu_{2}, \mu_{2}-\mu_{3}\), \(\mu_{3}-\mu_{4}, \mu_{4}-\left(\mu_{5}+\mu_{6}\right) / 2\), and \(\left(\mu_{1}+\mu_{2}+\cdots+\mu_{6}\right) / 6 .\) Here \(m=5 .\) Show that the lengths of the confidence intervals given by the results of Part (b) are shorter than the corresponding ones given by the method of Scheff茅 as described in the text. If \(m\) becomes sufficiently large, however, this is not the case.
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