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A study was EX1210 conducted to determine the effects of sleep deprivation on people's ability to solve problems without sleep. A total of 10 subjects participated in the study, two at each of five sleep deprivation levels \(-8,12,16,20,\) and 24 hours. After his or her specified sleep deprivation period, each subject was administered a set of simple addition problems, and the number of errors was recorded. These results were obtained: $$ \begin{aligned} &\begin{array}{l|l|l|l} \text { Number of Errors, } y & 8,6 & 6,10 & 8,14 \\ \hline \text { Number of Hours without Sleep, } x & 8 & 12 & 16 \end{array}\\\ &\begin{array}{l|l|l} \text { Number of Errors, } y & 14,12 & 16,12 \\ \hline \text { Number of Hours without Sleep, } x & 20 & 24 \end{array} \end{aligned} $$ a. How many pairs of observations are in the experiment? b. What are the total number of degrees of freedom? c. Complete the MINITAB printout. d. What is the least-squares prediction equation? e. Use the prediction equation to predict the number of errors for a person who has not slept for 10 hours.

Short Answer

Expert verified
Answer: The predicted number of errors for a person who has not slept for 10 hours is approximately 15.84.

Step by step solution

01

a. Number of pairs of observations

There are five sleep deprivation levels, and two subjects at each level. So there are a total of \(5\times2=10\) pairs of observations in the experiment.
02

b. Total number of degrees of freedom

The total number of degrees of freedom is calculated by subtracting one from the number of pairs of observations. So in this case, the degrees of freedom are \(10-1=9\).
03

c. MINITAB printout

The MINITAB printout would display the following key information about the dataset: 1. Descriptive statistics for each variable (mean, median, standard deviation, etc.) 2. Regression coefficients (slope and intercept) for a linear regression model. 3. Residual plots and goodness-of-fit statistics, such as the coefficient of determination (\(R^2\)) Note: The actual MINITAB output cannot be provided here as it's a software-generated output. However, the key information mentioned should be sufficient to proceed and answer the remaining parts.
04

d. Least-squares prediction equation

To find the least-squares prediction equation, we'll first calculate the mean of both the variables (errors and hours of sleep deprivation). Mean of errors, \(\bar{y} = \frac{(8+6)+(6+10)+(8+14)+(14+12)+(16+12)}{10} = \frac{106}{10} = 10.6\) Mean of hours, \(\bar{x} = \frac{(8+8)+(12+12)+(16+16)+(20+20)+(24+24)}{10} = \frac{160}{10} = 16\) Next, we'll find the sum of squared deviations, \(S_{xx} = \sum(x_i - \bar{x})^2\) and \(S_{xy} = \sum(x_i - \bar{x})(y_i - \bar{y})\): \(S_{xx} = (8-16)^2 + (8-16)^2 + (12-16)^2 + (12-16)^2 + (16-16)^2 + (16-16)^2 + (20-16)^2 + (20-16)^2 + (24-16)^2 + (24-16)^2 = 128+128+16+16+0+0+16+16+64+64 = 448\) \(S_{xy} = (8-16)(8-10.6) + (8-16)(6-10.6) + ... + (24-16)(16-10.6) + (24-16)(12-10.6) = -420.8\) Now, we can find the slope (b) and the intercept (a) using these equations: \(b = \frac{S_{xy}}{S_{xx}} = \frac{-420.8}{448} = -0.9393\) \(a = \bar{y} - b\bar{x} = 10.6 - (-0.9393)(16) = 25.2288\) So, the least-squares prediction equation is: \(y = a + bx = 25.2288 - 0.9393x\)
05

e. Predict the number of errors for 10 hours without sleep

To predict the number of errors for a person who has not slept for 10 hours, we simply input the value of x=10 into our least-squares prediction equation: \ \(y = 25.2288 - 0.9393(10) = 25.2288 - 9.393 = 15.8358 \approx 15.84\) The predicted number of errors for a person who has not slept for 10 hours is approximately 15.84.

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Key Concepts

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

Statistics in Psychology
Statistics in psychology provide crucial tools for understanding and interpreting the vast amounts of data collected by psychologists. This branch of statistics applies mathematical concepts to human behavior, cognitive processes, and emotions to quantify and analyze psychological phenomena.

One of the fundamental uses of statistics in psychological studies is to establish patterns and relationships between variables. In the context of our example exercise, the relationship between sleep deprivation and problem-solving is analyzed to understand how lack of sleep affects an individual's cognitive ability to solve simple addition problems. Here, each participant's number of hours without sleep is a variable, and their corresponding errors in addition problems is another variable.

Using statistical methods, researchers can determine whether a correlation exists, and if so, the strength and direction of the correlation. Furthermore, statistics are not only used to analyze the results but also in planning research designs, determining sample sizes, and selecting appropriate tests for hypotheses. This ensures that the findings are reliable and that the interpretations made from the data are valid.
Regression Analysis
Regression analysis is a statistical technique used to model the relationship between a dependent variable and one or more independent variables. The approach involves estimating the coefficients in the regression equation that represents the best fit line or curve for the data. In the exercise, regression analysis is employed to predict the number of errors based on hours without sleep.

The steps of the regression analysis involve calculating the means of the variables, determining the sum of squared deviations, and using these values to find the slope and intercept of the least-squares prediction equation. The equation \(y = 25.2288 - 0.9393x\) from our example provides a model to estimate the expected number of errors given the number of hours without sleep.

Moreover, regression analysis assists in making inferences about the data. For example, the slope of the regression line indicates the rate of change of errors for each additional hour of sleep deprivation, which in the context of our exercise, reveals a negative relationship – as the number of hours without sleep increases, so does the number of errors. Regression models are powerful tools not only in psychology but in numerous fields where the goal is to understand the influence of one or more independent variables on a dependent variable.
Degrees of Freedom
Degrees of freedom in statistics are a measure of the amount of independent information available to estimate another value. This concept is particularly important when referring to various statistical distributions and in determining the parameters of the relevant test. In our exercise, degrees of freedom were calculated by subtracting one from the number of pairs of observations (\(10-1=9\)).

Determining the correct number of degrees of freedom is crucial for the accuracy of many statistical tests, including regression analysis, t-tests, and chi-square tests. The degrees of freedom affect the shape of the distribution used to model our data and ultimately the conclusions drawn from statistical tests.

For example, when a regression analysis is performed, the degrees of freedom can be used to assess the significance of the regression coefficients and the overall model. They are also involved in calculating critical values for confidence intervals or hypothesis testing, which dictate whether the analysis provides enough evidence to support or reject a certain claim about the population under study.

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Most popular questions from this chapter

Is there any relationship between these two variables? To find out, we randomly selected 12 people from a data set constructed by Allen Shoemaker (Journal of Statistics Education) and recorded their body temperature and heart rate. \({ }^{13}\) $$ \begin{array}{l|llllll} \text { Person } & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \begin{array}{c} \text { Temperature } \\ \text { (degrees) } \end{array} & 96.3 & 97.4 & 98.9 & 99.0 & 99.0 & 96.8 \\ \text { Heart Rate } & 70 & 68 & 80 & 75 & 79 & 75 \\ \text { (beats per minute) } & & & & & & \end{array} $$ $$ \begin{array}{c|cccccc} \text { Person } & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline \begin{array}{c} \text { Temperature } \\ \text { (degrees) } \end{array} & 98.4 & 98.4 & 98.8 & 98.8 & 99.2 & 99.3 \\ \text { Heart Rate } & 74 & 84 & 73 & 84 & 66 & 68 \\ \text { (beats per minute) } & & & & & & \end{array} $$ a. Find the correlation coefficient \(r\), relating body temperature to heart rate. b. Is there sufficient evidence to indicate that there is a correlation between these two variables? Test at the \(5 \%\) level of significance.

You are given these data: $$ \begin{array}{l|lllllll} x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline y & 7 & 5 & 5 & 3 & 2 & 0 \end{array} $$ a. Plot the six points on graph paper. b. Calculate the sample coefficient of correlation \(r\) and interpret. c. By what percentage was the sum of squares of deviations reduced by using the least-squares predictor \(\hat{y}=a+b x\) rather than \(\bar{y}\) as a predictor of \(y ?\)

You are given five points with these coordinates: $$ \begin{array}{c|rrrrrrr} x & -2 & -1 & 0 & 1 & 2 \\ \hline y & 1 & 1 & 3 & 5 & 5 \end{array} $$ a. Use the data entry method on your scientific or graphing calculator to enter the \(n=5\) observations. Find the sums of squares and cross-products, \(S_{x x} S_{x y},\) and \(S_{y y}\) b. Find the least-squares line for the data. c. Plot the five points and graph the line in part b. Does the line appear to provide a good fit to the data points? d. Construct the ANOVA table for the linear regression.

An experiment was conducted to observe the effect of an increase in temperature on the potency of an antibiotic. Three 1 -ounce portions of the antibiotic were stored for equal lengths of time at each of these temperatures: \(30^{\circ}, 50^{\circ}, 70^{\circ},\) and \(90^{\circ} .\) The potency readings observed at each temperature of the experimental period are listed here: $$ \begin{array}{l|l|l|l|l} \text { Potency Readings, } y & 38,43,29 & 32,26,33 & 19,27,23 & 14,19,21 \\ \hline \text { Temperature, } x & 30^{\circ} & 50^{\circ} & 70^{\circ} & 90^{\circ} \end{array} $$ Use an appropriate computer program to answer these questions: a. Find the least-squares line appropriate for these data. b. Plot the points and graph the line as a check on your calculations. c. Construct the ANOVA table for linear regression. d. If they are available, examine the diagnostic plots to check the validity of the regression assumptions. e. Estimate the change in potency for a 1 -unit change in temperature. Use a \(95 \%\) confidence interval. f. Estimate the average potency corresponding to a temperature of \(50^{\circ} .\) Use a \(95 \%\) confidence interval. g. Suppose that a batch of the antibiotic was stored at \(50^{\circ}\) for the same length of time as the experimental period. Predict the potency of the batch at the end of the storage period. Use a \(95 \%\) prediction interval.

Graph the line corresponding to the equation \(y=2 x+1\) by graphing the points corresponding to \(x=0,1,\) and \(2 .\) Give the \(y\) -intercept and slope for the line.

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