/*! 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 72 The accompanying data on \(x=\) ... [FREE SOLUTION] | 91Ó°ÊÓ

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The accompanying data on \(x=\) student-teacher ratio is for a random sample of 20 high schools in Maine selected from a population of 85 high schools. The data are consistent with summary values for the state of Maine that appeared in an article in the Bangor Daily News (September \(22,2016,\) bangordailynews.com/2016/09/22/mainefocus/we-discovered-a-surprise-when-we- looked-deeper-into-our-survey-of-maine-principals/?ref=morelnmidcoast, retrieved May 2, 2017). The corresponding normal scores are also shown. $$ \begin{array}{|cc|} \hline \text { Student-Teacher Ratio }(x) & \text { Normal Score } \\ \hline 9.0 & -1.868 \\ 10.0 & -1.403 \\ 11.0 & -1.128 \\ 11.2 & -0.919 \\ 11.6 & -0.744 \\ 11.7 & -0.589 \\ 11.8 & -0.448 \\ 11.9 & -0.315 \\ 12.0 & -0.187 \\ 12.1 & -0.062 \\ 12.5 & 0.062 \\ 12.6 & 0.187 \\ 13.0 & 0.315 \\ 13.2 & 0.448 \\ 13.6 & 0.589 \\ 13.7 & 0.744 \\ 14.0 & 0.919 \\ 14.5 & 1.128 \\ 14.9 & 1.403 \\ 15.0 & 1.868 \\ \hline \end{array} $$ a. Construct a normal probability plot. b. Calculate the correlation coefficient for the (normal score, \(x\) ) pairs. Compare this value to the appropriate critical \(r\) value from Table 6.2 to determine if it is reasonable to think that the distribution of student-teacher ratios for high schools in Maine is approximately normal.

Short Answer

Expert verified
In summary, to determine if the distribution of student-teacher ratios for high schools in Maine is approximately normal, construct a normal probability plot and calculate the correlation coefficient (r) for the given data. Then, compare the calculated r value with the critical r value from Table 6.2. If the calculated r value is greater than or equal to the critical r value, it is reasonable to think that the distribution is approximately normal.

Step by step solution

01

Construct a normal probability plot

To construct the normal probability plot, we can plot the given normal scores on the horizontal axis and student-teacher ratios on the vertical axis. Each point on the plot will represent a pair (normal score, student-teacher ratio). You can do this using graph paper or graphing software.
02

Calculate the correlation coefficient (r) for the pairs

We're given the normal scores and student-teacher ratios in a table. To calculate the correlation coefficient, we'll first need to find the mean and standard deviation for both sets of data. 1. Compute the mean of normal scores (\( \bar{y} \)) and student-teacher ratios (\( \bar{x} \)) 2. Compute the standard deviation of normal scores (\( s_y \)) and student-teacher ratios (\( s_x \)) 3. Calculate the correlation coefficient using the formula: \(r = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum (x_i - \bar{x})^2}\sqrt{\sum (y_i - \bar{y})^2}}\)
03

Compare the calculated correlation coefficient (r) with the critical value

Once we have the correlation coefficient, we can compare it with the critical r value from Table 6.2. If the calculated r value is greater than or equal to the critical r value, we can conclude that it is reasonable to think that the distribution of student-teacher ratios for high schools in Maine is approximately normal. If the calculated r value is less than the critical r value, the distribution is not normal, and we cannot make any conclusions about the normality of the data. By following the steps above, you can calculate the correlation coefficient for the given data and compare it with the critical r value to determine if the distribution of student-teacher ratios for high schools in Maine is approximately normal.

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Key Concepts

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

Understanding Student-Teacher Ratio
The student-teacher ratio in schools is a simple but valuable indicator, reflecting the number of students for every teacher in an educational institution. A lower ratio often suggests that students might receive more personal attention and tailored instruction, whereas a higher ratio can imply less individualized attention. In the context of our data from Maine high schools, the ratios give insight into the educational environment and can be used to draw comparisons or correlations with educational outcomes.
When analyzing the significance of the student-teacher ratio, one can look at several factors. For instance, data on student-teacher ratios may correlate with student performance, availability of resources, and overall educational quality. Therefore, a thorough understanding of these ratios and their implications on education can be very helpful for policymakers, administrators, and educators who aim to improve the learning experience.
Deciphering the Correlation Coefficient
The correlation coefficient, denoted as 'r', quantifies the strength and direction of a linear relationship between two variables. It's a statistic that can take a value from -1 to +1. A value closer to +1 indicates a strong positive correlation, meaning as one variable increases, so does the other. Conversely, a value closer to -1 indicates a strong negative correlation, where one variable increases as the other decreases. A value around 0 suggests little to no linear relationship.
In the given exercise, calculating the correlation coefficient between normal scores and student-teacher ratios can reveal if the relationship between these two variables is linear and positive, linear and negative, or nonexistent. This calculation involves summing the products of each pair's deviations from their respective means and normalizing this sum by the product of both variables' standard deviations. The conclusion drawn from the correlation coefficient is crucial since it determines whether the assumption of normality in the student-teacher ratios can be reasonably accepted.
Analyzing Normal Distribution
Normal distribution analysis is a critical part of statistical data interpretation. It assumes that the data points are spread in a pattern where most of the observations cluster around the mean, creating a bell-shaped curve when plotted. This distribution is symmetrical, with the mean, median, and mode being equal.
In our case, a normal probability plot is used to assess whether the sample data on student-teacher ratios follow a normal distribution. On this plot, the data points should form roughly a straight line if the distribution is normal. Deviations from this line could indicate skewness or outliers in the data. By comparing the plot with the correlation coefficient, we can draw more conclusive evidence about the normalcy of our data. If the distribution is normal, this can streamline further statistical analyses, such as hypothesis testing or regression models, since many statistical methods assume normality of the data.

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Most popular questions from this chapter

A contractor is required by a county planning department to submit anywhere from one to five forms (depending on the nature of the project) when applying for a building permit. Let \(y\) be the number of forms required of the next applicant. Suppose the probability that \(y\) forms are required is known to be proportional to \(y ;\) that is, \(p(y)=\) \(k y\) for \(y=1, \ldots, 5\) a. What is the value of \(k ?\) (Hint: \(\Sigma p(y)=1 .)\) b. What is the probability that at most three forms are required? c. What is the probability that between two and four forms (inclusive) are required?

Determine each of the following areas under the standard normal (z) curve: a. To the left of -1.28 b. To the right of 1.28 c. Between -1 and 2 d. To the right of 0 e. To the right of -5 f. Between -1.6 and 2.5 g. To the left of 0.23

Suppose that \(20 \%\) of all homeowners in an earthquakeprone area of California are insured against earthquake damage. Four homeowners are selected at random. Define the random variable \(x\) as the number among the four who have earthquake insurance. a. Find the probability distribution of \(x\). (Hint: Let \(S\) denote a homeowner who has insurance and \(\mathrm{F}\) one who does not. Then one possible outcome is SFSS, with probability (0.2)(0.8)(0.2)(0.2) and associated \(x\) value of 3 . There are 15 other outcomes.) b. What is the most likely value of \(x ?\) c. What is the probability that at least two of the four selected homeowners have earthquake insurance?

Suppose that fuel efficiency (miles per gallon, mpg) for a particular car model under specified conditions is normally distributed with a mean value of \(30.0 \mathrm{mpg}\) and a standard deviation of \(1.2 \mathrm{mpg}\). a. What is the probability that the fuel efficiency for a randomly selected car of this model is between 29 and \(31 \mathrm{mpg}\) ? b. Would it surprise you to find that the efficiency of a randomly selected car of this model is less than \(25 \mathrm{mpg} ?\) c. If three cars of this model are randomly selected, what is the probability that each of the three have efficiencies exceeding \(32 \mathrm{mpg}\) ? d. Find a number \(x^{*}\) such that \(95 \%\) of all cars of this model have efficiencies exceeding \(x^{*}\) (i.e., \(P\left(x>x^{*}\right)=0.95\) ).

6.89 Suppose that \(20 \%\) of the 10,000 signatures on a certain recall petition are invalid. Would the number of invalid signatures in a sample of size 2000 have (approximately) a binomial distribution? Explain.

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