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Television is regarded by many as a prime culprit for the difficulty many students have in performing well in school. The article "The Impact of Athletics, Part-Time Employment, and Other Activities on Academic Achievement" (Journal of College Student Development [1992]: \(447-453\) ) reported that for a random sample of \(n=528\) college students, the sample correlation coefficient between time spent watching television \((x)\) and grade point average \((y)\) was \(r=-.26\). a. Does this suggest that there is a negative correlation between these two variables in the population from which the 528 students were selected? Use a test with significance level .01. b. If \(y\) were regressed on \(x\), would the regression explain a substantial percentage of the observed variation in grade point average? Explain your reasoning.

Short Answer

Expert verified
a. Yes, we reject the null hypothesis and suggest with 99% confidence that there's a negative correlation in the population from which students were sampled. b. Only 6.76% of the variability in GPA can be explained by the time spent watching TV, so the regression would not explain a substantial variation in GPA.

Step by step solution

01

State the Hypothesis

The null hypothesis \(H_0\) is that the population correlation coefficient \(\rho = 0\), meaning there is no correlation. The alternative hypothesis \(H_a\) is that the population correlation coefficient \(\rho != 0\), i.e., there is some correlation. Here, we're particularly interested in whether there's a negative correlation, so we could also formulate the alternative hypothesis as \(\rho < 0\).
02

Calculate Test Statistic

The test statistic in this case is the correlation coefficient \(r\), which is \(-.26\). Since we're dealing with a large sample size (\(n = 528 > 30\)), the sampling distribution of the correlation coefficient can be approximated as normal.
03

Identify the Critical Value

We need to compare our test statistic with the critical value to decide whether to reject the null hypothesis. As our significance level is .01, and this is a one-tailed test, we look up the critical value for z in the standard normal distribution (Z-table) corresponding to a cumulative probability of .99. The critical value is approximately 2.33.
04

Decision

Our correlation coefficient of \(-.26\) is less than our critical value of 2.33, therefore we have enough evidence to reject the null hypothesis and claim that the correlation in the population is indeed negative. Thus, there appears to be a negative correlation between TV watching time and grade point average in the population.
05

Answering Part B

To answer whether regressing y (GPA) on x (TV watching time) would explain a substantial amount of the variation, we need to calculate the coefficient of determination. This is simply the square of the correlation coefficient, \(r^2 = (-0.26)^2 = 0.0676\). This tells us that only about 6.76% of the variation in GPA can be explained by the TV watching time. Therefore, this regression would not explain a substantial proportion of the observed variation in GPA.

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Key Concepts

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

Hypothesis Testing
In statistics, hypothesis testing is a fundamental process used to determine if a certain effect or relationship exists in a population based on sample data. The reasoning is quite intuitive: we start with an assumption (the null hypothesis, usually symbolized as \(H_0\)) that there is no effect or relationship. Then, we calculate how likely it is to observe our sample data if this null hypothesis were true. If it's very unlikely, we conclude that our initial assumption might be wrong, and there is evidence against it.

In the context of the correlation coefficient example, we have a null hypothesis stating no correlation between television watching and grade point average (\(H_0: \rho = 0\)) and an alternative hypothesis suggesting a negative correlation (\(H_a: \rho < 0\)). By comparing the calculated correlation coefficient in the sample to the critical value derived from statistical tables, we can determine if our evidence strongly contradicts the null hypothesis, leading us to either reject or fail to reject \(H_0\). It's like a courtroom trial for the assumed relationship – and just as in a trial, our decision to reject is based on the threshold of 'reasonable doubt', known as the significance level.
Regression Analysis
Regression analysis is a powerful statistical tool that models the relationship between a dependent variable and one or more independent variables. It’s akin to figuring out which pieces, and to what extent, contribute to the completion of a puzzle.

In our example, regression analysis would help us understand how much time spent watching television (independent variable) impacts grade point average (dependent variable). With the correlation hinting at a negative relationship, regression could quantify this relationship, perhaps leading to an equation that predicts GPA based on TV watching time. However, just like a single piece doesn’t make a puzzle, a regression with a low explanatory power might indicate that the model (or the piece) isn’t quite the right fit for our data puzzle, and we need to look for additional pieces (variables) to complete it.
Coefficient of Determination
When it comes to understanding how well our model fits the data, the coefficient of determination, symbolized as \(R^2\), is our go-to statistic. It quantifies the proportion of the variance in the dependent variable that's predictable from the independent variable(s). Think of it as a score that tells you how much of the story is explained by the factors you've included in your model.

In the example provided, we calculate the coefficient of determination by squaring the correlation coefficient (\(r^2 = (-0.26)^2 = 0.0676\)). With an \(R^2\) of merely 6.76%, we deduce that watching TV accounts for just under 7% of the variation in GPA. This is a low score, signifying that our model – if TV watching were the only variable considered – leaves a lot of the story untold. Other factors could also be influencing GPA, and they’re not captured by this simple model.
Sample Size
The sample size, denoted \(n\), is the number of observations in a study and is critical to the reliability of statistical analyses. It’s like the audience for a play; the bigger the audience, the more likely you are to have a diverse group and the more confident you can be in the overall reaction reflecting the general public’s opinion.

In hypothesis testing, the sample size affects the precision of our estimates and the strength of our conclusions. With a large sample size like in our example (\(n = 528\)), we have more information, which generally leads to more precise estimates of the population parameters and more power to detect actual effects. That’s why when the sample size is large enough, statistical procedures can be applied with greater confidence in their results.
Critical Value
The critical value acts as the threshold or cut-off point in hypothesis testing. Just as a high-jumper needs to clear the bar to be successful, our test statistic must 'clear' the critical value to provide compelling evidence against the null hypothesis.

It's selected based on the chosen significance level (the probability of mistakenly rejecting a true null hypothesis) and the nature of the test (one-tailed or two-tailed). In our exercise, since we are considering a one-tailed test with a significance level of 0.01, we look for the value in the standard normal distribution table that corresponds to 99% confidence (the point where only 1% of the distribution lies beyond). The larger the magnitude of the critical value, the more stringent the test, and the harder it is to reject the null hypothesis. When our test statistic (in this case, the sample correlation coefficient) is beyond this critical value, we have significant statistical evidence – or in our high-jump analogy, the athlete has cleared the bar and made a new record.

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Most popular questions from this chapter

The article "Technology, Productivity, and Industry Structure" (Technological Forecasting and Social Change [1983]: \(1-13\) ) included the accompanying data on \(x=\) research and development expenditure and \(y=\) growth rate for eight different industries. $$ \begin{array}{rrrrrrrrr} x & 2024 & 5038 & 905 & 3572 & 1157 & 327 & 378 & 191 \\ y & 1.90 & 3.96 & 2.44 & 0.88 & 0.37 & -0.90 & 0.49 & 1.01 \end{array} $$ a. Would a simple linear regression model provide useful information for predicting growth rate from research and development expenditure? Use a .05 level of significance. b. Use a \(90 \%\) confidence interval to estimate the average change in growth rate associated with a 1 -unit increase in expenditure. Interpret the resulting interval.

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Some straightforward but slightly tedious algebra shows that $$ \text { SSResid }=\left(1-r^{2}\right) \sum(y-\bar{y})^{2} $$ from which it follows that $$ s_{e}=\sqrt{\frac{n-1}{n-2}}\left(\sqrt{1-r^{2}}\right) s_{y} $$ Unless \(n\) is quite small, \((n-1) /(n-2) \approx 1\), so $$ s_{e} \approx\left(\sqrt{1-r^{2}}\right) s_{y} $$ a. For what value of \(r\) is \(s_{e}\) as large as \(s_{y}\) ? What is the equation of the least-squares line in this case? b. For what values of \(r\) will \(s_{e}\) be much smaller than \(s_{y} ?\)

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