/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 Consider the accompanying data o... [FREE SOLUTION] | 91影视

91影视

Consider the accompanying data on \(x=\) research and development expenditure (thousands of dollars) and \(y=\) growth rate (\% per year) for eight different industries. \(\begin{array}{lllllllll}x & 2024 & 5038 & 905 & 3572 & 1157 & 327 & 378 & 191\end{array}\) \(\begin{array}{ccccccccc}y & 1.90 & 3.96 & 2.44 & 0.88 & 0.37 & -0.90 & 0.49 & 1.01\end{array}\) 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 \(\$ 1000\) increase in expenditure. Interpret the resulting interval.

Short Answer

Expert verified
Without the actual computation, a definitive answer cannot be given. However, if the correlation coefficient is found to be statistically significant, then a linear regression model can provide useful information for predicting growth rate based on expenditure. The 90% confidence interval for the slope will give the range of possible values for the average change in growth rate associated with a $1000 increase in R&D expenditure, which can be interpreted accordingly.

Step by step solution

01

Calculate the Correlation Co-efficient

Firstly, the correlation coefficient (\(r\)) needs to be calculated to determine if a linear relationship exists between \(x\) and \(y\). This can be done using a statistical software or calculator.
02

Perform Linear Regression Analysis

Once \(r\) has been found, use the data to create a linear regression model. With this model, parameters, such as the slope and intercept, are calculated. These will be used for prediction. Interpret the coefficients obtained to make valid conclusions for part b of the exercise.
03

Test the Significance of the Correlation Coefficient

With \(r\) and the model parameters calculated, perform a hypothesis test to determine if the correlation coefficient is significantly different from zero at a \(0.05\) significance level. If significant, a linear regression model can provide useful information for predicting growth rate from expenditure.
04

Construct a 90% Confidence Interval for the Slope

Use the calculated slope and its standard error to construct a 90% confidence interval. This interval will estimate the average change in growth rate associated with a $1000 increase in R&D expenditure.
05

Interpret the Resulting Interval

The interval obtained in Step 4 represents the range of values where the true slope (average change in growth rate per $1000 increase in expenditure) falls, with 90% certainty. Interpret this interval in the context of the problem.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Correlation Coefficient
The correlation coefficient, commonly denoted by the symbol \( r \), offers a measure of the strength and direction of the linear relationship between two variables. Think of it as a numerical summary that ranges between -1 and 1.

An \( r \) value closer to 1 indicates a strong positive correlation, meaning as one variable increases, so does the other. Conversely, an \( r \) value closer to -1 suggests a strong negative correlation, where one variable鈥檚 increase corresponds to the other鈥檚 decrease. When \( r \) is around 0, it implies little to no linear relationship between the two variables.

In the context of the exercise, calculating \( r \) between research and development expenditure and the growth rate will help us understand whether higher investments in R&D are associated with higher growth rates within industries. This insight is crucial as it serves as the foundation for building a linear regression model that could reliably predict future growth rates based on R&D expenditure.
Hypothesis Test
Hypothesis testing is a statistical method that allows us to make decisions about a population parameter based on a sample statistic. In the case of a simple linear regression, the null hypothesis \( (H_0) \) often states that there is no relationship between the two variables, which would mean our correlation coefficient \( r \) equals zero.

To test this hypothesis, we generally set a significance level (like \( 0.05 \)), which determines the probability of rejecting the null hypothesis when it鈥檚 actually true (Type I error). If the p-value obtained from the hypothesis test is less than the significance level, we reject the null hypothesis, indicating there is enough evidence to suggest a statistically significant relationship between the variables.

Applying this to our exercise, we use a hypothesis test to confirm whether the correlation between R&D expenditure and growth rate is significant, and not just due to random chance. If significant, it supports the decision to use a linear regression model for prediction.
Confidence Interval
A confidence interval gives a range of values within which we can expect a population parameter to lie, within a certain level of confidence. It鈥檚 like saying, 鈥淲e are \( 90\% \) (or whatever confidence level we鈥檙e working with) certain that the true value is somewhere between this lower and upper bound.鈥

When we construct a confidence interval for the slope of a linear regression, it helps us understand the potential change in the response variable for a unit change in the predictor variable. In the exercise, the slope represents the expected change in growth rate for each \(1000 increase in R&D expenditure.

The confidence interval from part b of the exercise thus offers us a range that we are \( 90\% \) confident contains the true average change in growth rate per \)1000 of R&D expenditure, allowing us to make more informed business decisions based on this estimated interval.
Research and Development Expenditure
Research and Development (R&D) expenditure is a critical metric for industries, as it signifies investment in innovation and future capabilities. It鈥檚 an input variable we often want to examine to see its effect on various output variables, such as growth rate, productivity, or profitability.

In simple linear regression, we look at how R&D expenditure (independent variable) influences the growth rate (dependent variable). We expect that increased R&D spending would correlate with greater growth rates, under the assumption that investment in R&D drives innovation and, consequently, growth.

It's crucial to emphasize that although R&D expenditure is a valuable factor, its interpretation in the regression model needs to be considered alongside other potential factors affecting growth which may not be included in a simple linear regression model.
Growth Rate Prediction
Growth rate prediction involves estimating the rate at which a company or industry is expected to expand in a given period. It's an outcome variable that can be predicted through various statistical models, one of which is the linear regression model.

In our case, after determining there is a significant relationship between R&D expenditure and growth rate, we can use the regression model to predict future growth. The model generates an equation of the form \( y = mx + b \), where \( y \) is the predicted growth rate, \( m \) is the slope indicating the rate of change, and \( b \) is the intercept, representing the growth rate when expenditure is zero.

With this model, companies can forecast growth rates based on their R&D investment strategies, which is extremely beneficial for planning and resource allocation. Nevertheless, accuracy of these predictions depends on the reliability of the model, which is influenced by the quality of data, the strength of the correlation, and the suitability of linear regression to the data at hand.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The authors of the paper studied the relationship between childhood environmental lead exposure and a measure of brain volume change in a particular region of the brain. Data were given for \(x=\) mean childhood blood lead level \((\mu \mathrm{g} / \mathrm{dL})\) and \(y=\) brain volume change \((\mathrm{BVC}\), in percent \() .\) A subset of data read from a graph that appeared in the paper was used to produce the accompanying Minitab output. Carry out a hypothesis test to decide if there is convincing evidence of a useful linear relationship between \(x\) and \(y\). Assume that the basic assumptions of the simple linear regression model are reasonably met.

Carbon aerosols have been identified as a contributing factor in a number of air quality problems. In a chemical analysis of diesel engine exhaust, \(x=\) mass \(\left(\mu \mathrm{g} / \mathrm{cm}^{2}\right)\) and \(y=\) elemental carbon \(\left(\mu \mathrm{g} / \mathrm{cm}^{2}\right)\) were recorded. The estimated regression line for this data set is \(\hat{y}=31+.737 x\). The accompanying table gives the observed \(x\) and \(y\) values and the corresponding standardized residuals. \(\begin{array}{lrrrrr}x & 164.2 & 156.9 & 109.8 & 111.4 & 87.0 \\\ y & 181 & 156 & 115 & 132 & 96 \\ \text { St. resid. } & 2.52 & 0.82 & 0.27 & 1.64 & 0.08 \\ x & 161.8 & 230.9 & 106.5 & 97.6 & 79.7 \\ y & 170 & 193 & 110 & 94 & 77 \\ \text { St. resid. } & 1.72 & -0.73 & 0.05 & -0.77 & -1.11 \\\ x & 118.7 & 248.8 & 102.4 & 64.2 & 89.4 \\ y & 106 & 204 & 98 & 76 & 89 \\\ \text { St. resid. } & -1.07 & -0.95 & -0.73 & -0.20 & -0.68 \\ x & 108.1 & 89.4 & 76.4 & 131.7 & 100.8 \\ y & 102 & 91 & 97 & 128 & 88 \\ \text { St. resid. } & -0.75 & -0.51 & 0.85 & 0.00 & -1.49\end{array}\) \(\begin{array}{lllll}78.9 & 387.8 & 135.0 & 82.9 & 117.9\end{array}\) a. Construct a standardized residual plot. Are there any unusually large residuals? Do you think that there are any influential observations? b. Is there any pattern in the standardized residual plot that would indicate that the simple linear regression model is not appropriate? c. Based on your plot in Part (a), do you think that it is reasonable to assume that the variance of \(y\) is the same at each \(x\) value? Explain.

The authors of the paper studied a number of variables they thought might be related to bone mineral density (BMD). The accompanying data on \(x=\) weight at age 13 and \(y=\) bone mineral density at age 27 are consistent with summary quantities for women given in the paper. A simple linear regression model was used to describe the relationship between weight at age 13 and \(\mathrm{BMD}\) at age 27\. For this data: $$ \begin{array}{lll} a=0.558 & b=0.009 & n=15 \\ \mathrm{SSTo}=0.356 & \text { SSResid }=0.313 & \end{array} $$ a. What percentage of observed variation in \(\mathrm{BMD}\) at age 27 can be explained by the simple linear regression model? b. Give a point estimate of \(\sigma\) and interpret this estimate. c. Give an estimate of the average change in BMD associated with a \(1 \mathrm{~kg}\) increase in weight at age 13 . d. Compute a point estimate of the mean BMD at age 27 for women whose age 13 weight was \(60 \mathrm{~kg}\).

According to the size of a female salamander's snout is correlated with the number of eggs in her clutch. The following data are consistent with summary quantities reported in the article. Partial Minitab output is also included. \(\begin{array}{lrrrrr}\text { Snout-Vent Length } & 32 & 53 & 53 & 53 & 54 \\\ \text { Clutch Size } & 45 & 215 & 160 & 170 & 190 \\ \text { Snout-Vent Length } & 57 & 57 & 58 & 58 & 59 \\ \text { Clutch Size } & 200 & 270 & 175 & 245 & 215 \\ \text { Snout-Vent Length } & 63 & 63 & 64 & 67 & \\ \text { Clutch Size } & 170 & 240 & 245 & 280 & \end{array}\) 2 The regression equation is \(\begin{aligned}&Y=-133+5.92 x \\\&\text { Predictor } & \text { Coef } & \text { StDev } & T & P \\\&\text { Constant } & -133.02 & 64.30 & 2.07 & 0.061 \\\&x & 5.919 & 1.127 & 5.25 & 0.000 \\\&s=33.90 & \text { R-Sq }=69.7 \% & R-S q(a d j)=67.2 \%\end{aligned}\) Additional summary statistics are \(n=14 \quad \bar{x}=56.5 \quad \bar{y}=201.4\) \(\sum x^{2}=45,958 \quad \sum y^{2}=613,550 \quad \sum x y=164,969\) a. What is the equation of the regression line for predicting clutch size based on snout-vent length? b. What is the value of the estimated standard deviation of \(b\) ? c. Is there sufficient evidence to conclude that the slope of the population line is positive? d. Predict the clutch size for a salamander with a snoutvent length of 65 using a \(95 \%\) interval. e. Predict the clutch size for a salamander with a snoutvent length of 105 using a \(90 \%\) interval.

A sample of small cars was selected, and the values of \(x=\) horsepower and \(y=\) fuel efficiency \((\mathrm{mpg})\) were determined for each car. Fitting the simple linear regression model gave the estimated regression equation \(\hat{y}=44.0-.150 x .\) a. How would you interpret \(b=-.150\) ? b. Substituting \(x=100\) gives \(\hat{y}=29.0\). Give two different interpretations of this number. c. What happens if you predict efficiency for a car with a 300-horsepower engine? Why do you think this has occurred? d. Interpret \(r^{2}=0.680\) in the context of this problem. e. Interpret \(s_{e}=3.0\) in the context of this problem.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.