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The employee relations manager of a large company was concemed that raises given to employees during a recent period might not have been based strictly on objective performance criteria. A sample of \(n=20\) employees was selected, and the values of \(x\), a quantitative measure of productivity, and \(y\), the percentage salary increase, were determined for each one. A computer package was used to fit the simple linear regression model, and the resulting output gave the \(P\) -value \(=.0076\) for the model utility test. Does the percentage raise appear to be linearly related to productivity? Explain.

Short Answer

Expert verified
Yes, the percentage raise appears to be linearly related to productivity as deduced from the fact that the P-value for the model utility test (0.0076) is less than common significance levels (0.05, 0.01). This indicates there is sufficient evidence to reject the null hypothesis and accept the alternative hypothesis that there is a significant linear relationship between productivity and percentage salary increase.

Step by step solution

01

Understand the Problem

We have to determine whether the percentage salary increase is linearly related to productivity. This is gauged by a test where the null hypothesis is that there is no relationship i.e., the slope of the line in the simple linear regression model is zero. The alternative hypothesis is that there is a relationship i.e., the slope is not equal to zero.
02

Understand the Importance of the P-value

A P-value is the probability of obtaining a result as extreme as, or more extreme than, the observed result, under the null hypothesis. Small P-values (typically, P-value<0.05 or 0.01) would provide strong evidence against the null hypothesis hence we reject the null hypothesis. Conversely, if P-value>0.05 or 0.01, there is weak evidence against the null hypothesis hence we fail to reject the null hypothesis.
03

Compare the Given P-value with the Significance Levels

The given P-value is 0.0076. Comparing this value with the typical significance levels 0.05 and 0.01, we find that 0.0076 is less than both 0.01 and 0.05.
04

Make a Conclusion

Since the P-value (0.0076) is less than the typical significance levels of 0.05 and 0.01, we reject the null hypothesis and conclude that there is a significant relationship between productivity and the percentage salary increase. This means it appears the percentage raise is linearly related to productivity.

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Key Concepts

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

P-value
In statistics, the P-value is a fundamental concept that represents the probability of obtaining the observed results, or more extreme ones, if the null hypothesis were actually true. Imagine flipping a coin and wanting to know if it's biased; you'd flip it multiple times and calculate the probability of getting the observed mix of heads and tails assuming the coin is fair. That probability is similar to the P-value in statistical tests.

In the context of simple linear regression, a low P-value indicates that the observed correlation between two variables is unlikely to have occurred by random chance. This typically leads researchers to reject the null hypothesis. A common threshold for 'low' is usually 0.05 or even 0.01, meaning there's less than a 5% or 1% chance respectively of the results occurring if the null hypothesis were true. The exercise mentioned a P-value of 0.0076, which is below 0.05 and 0.01, suggesting a statistically significant relationship between productivity and salary increases.
Productivity Measurement
Productivity measurement is crucial in assessing how efficiently resources like labor or capital are utilized to produce outputs. In a business setting, measuring an employee's productivity can involve various metrics, such as the quantity of goods produced, the quality of work, or revenue generated. The exercise references a quantitative measure of productivity, denoted as x, which was used in evaluating whether salary increases should be based on this measure.

When productivity is quantified, it can be correlated with other factors, such as salary increases (denoted as y), using tools like regression analysis to see if higher productivity effectively translates to higher pay. Understanding these relationships can help companies in making data-driven decisions regarding compensation and performance incentives.
Regression Model Utility Test
The regression model utility test, also known as the F-test, is used in simple linear regression to determine whether the model provides a better fit to the data than a model with no independent variables. In other words, it tests whether the explanatory variables in the regression model significantly predict the outcome variable.

In our exercise, the test was done to evaluate if the increases in employees' salaries were based on their productivity rather than being arbitrary. This F-test gives us a P-value which we use to decide if we can accept or reject the null hypothesis about the utility of our regression model. With a low P-value in this particular instance, we have evidence that the linear regression model is indeed useful in explaining the relationship between productivity and salary raises.
Null Hypothesis
The null hypothesis serves as a starting assumption for statistical significance testing. It commonly posits that there is no effect or no difference, and in the case of regression, that there is no relationship between the independent and dependent variables. The null hypothesis for our exercise would be that there is no linear relationship between productivity and percentage salary increase.

The burden of proof lies in demonstrating that the null hypothesis can be rejected, which would suggest there is indeed an effect or a difference to be considered. When the P-value is smaller than the chosen significance level, it undermines the credibility of the null hypothesis, leading statisticians to conclude that it should be rejected, favoring the alternative hypothesis that proposes some form of relationship or effect.

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Most popular questions from this chapter

'The article "Photocharge Effects in Dye Sensitized \(\mathrm{Ag}[\mathrm{Br}, \mathrm{I}]\) Emulsions at Millisecond Range Exposures" (Photographic Science and Engineering [1981]: \(138-144\) ) gave the accompanying data on \(x=\%\) light absorption and \(y=\) peak photovoltage. \(\begin{array}{rrrrrrrrrr}x & 4.0 & 8.7 & 12.7 & 19.1 & 21.4 & 24.6 & 28.9 & 29.8 & 30.5 \\ y & 0.12 & 0.28 & 0.55 & 0.68 & 0.85 & 1.02 & 1.15 & 1.34 & 1.29\end{array}\) \(\sum x=179.7 \quad \sum x^{2}=4334.41\) \(\sum y=7.28 \quad \sum y^{2}=7.4028 \quad \sum x y=178.683\) a. Construct a scatterplot of the data. What does it suggest? b. Assuming that the simple linear regression model is appropriate, obtain the equation of the estimated regression line. c. How much of the observed variation in peak photovoltage can be explained by the model relationship? d. Predict peak photovoltage when percent absorption is 19.1, and compute the value of the corresponding residual. e. The authors claimed that there is a useful linear relationship between the two variables. Do you agree? Carry out a formal test. f. Give an estimate of the average change in peak photovoltage associated with a \(1 \%\) increase in light absorption. Your estimate should convey information about the precision of estimation. g. Give an estimate of true average peak photovoltage when percentage of light absorption is 20 , and do so in a way that conveys information about precision.

An investigation of the relationship between traffic flow \(x\) (thousands of cars per \(24 \mathrm{hr}\) ) and lead content \(y\) of bark on trees near the highway (mg/g dry weight) yielded the accompanying data. A simple linear regression model was fit, and the resulting estimated regression line was \(\hat{y}=28.7+33.3 x .\) Both residuals and standardized residuals are also given. \(\begin{array}{lrrrrr}\text { iduals are also given. } & & & & \\ x & 8.3 & 8.3 & 12.1 & 12.1 & 17.0 \\ y & 227 & 312 & 362 & 521 & 640 \\ \text { Residual } & -78.1 & 6.9 & -69.6 & 89.4 & 45.3 \\ \text { St. resid. } & -0.99 & 0.09 & -0.81 & 1.04 & 0.51\end{array}\) \(\begin{array}{lrrrrr}x & 17.0 & 17.0 & 24.3 & 24.3 & 24.3 \\ y & 539 & 728 & 945 & 738 & 759 \\ \text { Residual } & -55.7 & 133.3 & 107.2 & -99.8 & -78.8 \\\ \text { St. resid. } & -0.63 & 1.51 & 1.35 & -1.25 & -0.99\end{array}\) a. Plot the \((x\), residual \()\) pairs. Does the resulting plot suggest that a simple linear regression model is an appropriate choice? Explain your reasoning. b. Construct a standardized residual plot. Does the plot differ significantly in general appearance from the plot in Part (a)?

Are workers less likely to quit their jobs when wages are high than when they are low? The paper "Investigating the Causal Relationship Between Quits and Wages: An Exercise in Comparative Dynamics" (Economic Inquiry [1986]: \(61-83\) ) gave data on \(x=\) average hourly wage and \(y=\) quit rate for a sample of industries. These data were used to produce the accompanying MINITAB output The regression equation is quit rate \(=4.86-0.347\) wage Predictor Constant wage \(\begin{array}{rrrr}\text { Coef } & \text { Stdev } & \text { t-ratio } & p \\ 4.8615 & 0.5201 & 9.35 & 0.000 \\ 0.34655 & 0.05866 & 5.91 & 0.000\end{array}\) \(\begin{array}{lll}0.4862 & \mathrm{R}-\mathrm{sq}=72.9 \% & \mathrm{R}-\mathrm{sq}(\mathrm{ad}) & =70.8 \%\end{array}\) Analysis of Variance \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \\ \text { Regression } & 1 & 8.2507 & 8.2507 & 34.90 & 0.000 \\ \text { Error } & 13 & 3.0733 & 0.2364 & & \\ \text { Total } & 14 & 11.3240 & & & \end{array}\) a. Based on the given \(P\) -value, does there appear to be a useful linear relationship between average wage and quit rate? Explain your reasoning. b. Calculate an estimate of the average change in quit rate associated with a \(\$ 1\) increase in average hourly wage, and do so in a way that conveys information about the precision and reliability of the estimate.

The article "Effect of Temperature on the pH of Skim Milk" (Journal of Dairy Research [1988]: 277- 280) reported on a study involving \(x=\) temperature \(\left({ }^{\circ} \mathrm{C}\right)\) under specified experimental conditions and \(y=\) milk \(\mathrm{pH}\). The accompanying data (read from a graph) are a representative subset of that which appeared in the article: \(\begin{array}{rrrrrrrrr}x & 4 & 4 & 24 & 24 & 25 & 38 & 38 & 40 \\ y & 6.85 & 6.79 & 6.63 & 6.65 & 6.72 & 6.62 & 6.57 & 6.52\end{array}\) $$ \begin{array}{lrrrrrrrr} x & 45 & 50 & 55 & 56 & 60 & 67 & 70 & 78 \\ y & 6.50 & 6.48 & 6.42 & 6.41 & 6.38 & 6.34 & 6.32 & 6.34 \\ \sum x=678 & \sum y=104.54 & \sum x^{2}=36,056 & \\ \sum y^{2}=683.4470 & & \sum x y=4376.36 & & \end{array} $$ Do these data strongly suggest that there is a negative linear relationship between temperature and \(\mathrm{pH}\) ? State and test the relevant hypotheses using a significance level of \(.01\).

Occasionally an investigator may wish to compute a confidence interval for \(\alpha\), the \(y\) intercept of the true regression line, or test hypotheses about \(\alpha .\) The estimated \(y\) intercept is simply the height of the estimated line when \(x=0\), since \(a+b(0)=a .\) This implies that \(s_{a}\) the estimated standard deviation of the statistic \(a\), results from substituting \(x^{\prime \prime}=0\) in the formula for \(s_{a+b x^{+}} .\) The desired confidence interval is then \(a \pm(t\) critical value \() s_{a}\) \(-\) and a test statistic is 1 $$ t=\frac{a-\text { hypothesized value }}{s_{a}} $$ a. The article "Comparison of Winter-Nocturnal Geostationary Satellite Infrared-Surface Temperature with Shelter-Height Temperature in Florida" (Remote Sensing of the Emvironment \([1983]: 313-327\) ) used the simple linear regression model to relate surface temperature as measured by a satellite \((y)\) to actual air temperature \((x)\) as determined from a thermocouple placed on a traversing vehicle. Selected data are given (read from a scatterplot in the article). \(\begin{array}{rrrrrrrr}x & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ y & -3.9 & -2.1 & -2.0 & -1.2 & 0.0 & 1.9 & 0.6 \\ x & 5 & 6 & 7 & & & & \\ y & 2.1 & 1.2 & 3.0 & & & & \end{array}\) Estimate the true regression line. b. Compute the estimated standard deviation \(s_{a}\). Carry out a test at level of significance \(.05\) to see whether the \(y\) intercept of the true regression line differs from zero. c. Compute a \(95 \%\) confidence interval for \(\alpha\). Does the result indicate that \(\alpha=0\) is plausible? Explain.

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