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Here is a sample of ACT scores (average of the Math, English, Social Science, and Natural Science scores) for students taking college freshman calculus: \(\begin{array}{lllllll}24.00 & 28.00 & 27.75 & 27.00 & 24.25 & 23.50 & 26.25 \\\ 24.00 & 25.00 & 30.00 & 23.25 & 26.25 & 21.50 & 26.00 \\ 28.00 & 24.50 & 22.50 & 28.25 & 21.25 & 19.75 & \end{array}\) a. Using an appropriate graph, see if it is plausible that the observations were selected from a normal distribution. b. Calculate a two-sided \(95 \%\) confidence interval for the population mean. c. The university ACT average for entering freshmen that year was about 21. Are the calculus students better than average, as measured by the ACT?

Short Answer

Expert verified
Yes, the calculus students are better than average, as the interval [23.854, 26.351] is above the average of 21.

Step by step solution

01

Organize Data

First, arrange and list the sample ACT scores given: 24.00, 28.00, 27.75, 27.00, 24.25, 23.50, 26.25, 24.00, 25.00, 30.00, 23.25, 26.25, 21.50, 26.00, 28.00, 24.50, 22.50, 28.25, 21.25, 19.75.
02

Generate a Normal Probability Plot

Using statistical software or graphing tools, plot the ordered ACT scores on the y-axis against the theoretical quantiles of a normal distribution on the x-axis. This will help visually assess normality. Check if the points lie approximately on a straight line.
03

Compute Sample Mean and Standard Deviation

Calculate the sample mean (\(\bar{x}\)) and standard deviation (\(s\)). For this dataset: Mean \(\bar{x} = 25.1025\), Standard Deviation \(s = 2.676\).
04

Determine Sample Size

Count the number of observations in the sample to determine the sample size \(n\), which in this case is 20.
05

Calculate the Standard Error

Find the standard error (SE) of the sample mean using the formula: \(SE = \frac{s}{\sqrt{n}}\). For this data, \(SE = \frac{2.676}{\sqrt{20}} = 0.5985\).
06

Determine the Critical t-Value

For a 95% confidence interval and \(n-1 = 19\) degrees of freedom, look up the critical t-value (\(t^*\)) from a t-distribution table. \(t^* \approx 2.093\).
07

Calculate 95% Confidence Interval

Use the formula for the confidence interval: \(\bar{x} \pm t^* \cdot SE\). Calculate this as: \(25.1025 \pm 2.093 \times 0.5985\). The interval is \([23.854, 26.351]\).
08

Compare Against Population Mean

Compare the calculated confidence interval with the university's average ACT score of 21. Since 21 is not within the confidence interval \([23.854, 26.351]\), it suggests that the students are likely performing better than the average.

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Key Concepts

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

Normal Distribution
The concept of a normal distribution plays a crucial role in statistics, often used to model real-world data. In essence, a normal distribution is a symmetrical, bell-shaped curve representing the distribution of a dataset. The property of symmetry means that data tends to cluster around a central mean value, with data points becoming scarcer as they move away from the mean in either direction. This distribution is characterized by its mean (\(\mu\)) and standard deviation (\(\sigma\)).
  • The mean represents the central point of the data, providing the highest probability density.
  • The standard deviation measures the spread or dispersion of the data around the mean.
Visual checks, like a normal probability plot, help determine if data points follow this pattern. If most data points form a straight line, it's plausible they're from a normal distribution. This visual test is essential before conducting statistical analyses relying on normal distribution assumptions, ensuring validity and accuracy of results.
Confidence Interval
A confidence interval in statistics provides an estimated range of values, which is likely to contain the population parameter with a specified level of confidence. Here, we've calculated a 95% confidence interval for the population mean of the ACT scores.
The key idea is that if we were to take multiple samples, approximately 95% of the intervals calculated from these samples would contain the true population mean. This gives us a measure of certainty about where the actual mean lies.
  • The 95% confidence interval is calculated using the formula: \[\bar{x} \pm t^* \times SE\].
  • Where \(\bar{x}\) is the sample mean, \(t^*\) is the critical t-value, and \(SE\) is the standard error.
For the ACT scores in our sample, the confidence interval was determined to be \([23.854, 26.351]\). This interval suggests that we are 95% confident the true mean ACT score of the population is within this range. Importantly, the calculated interval does not include the university's overall mean of 21, indicating statistical significance regarding the performance level of calculus students.
Sample Mean
The sample mean is a fundamental statistic that provides a single value representing the average of a dataset. It’s essential for estimating the central tendency of the observed data points. We calculate the sample mean by summing all observed values and dividing by the number of observations.
In the ACT score scenario, the formula for the sample mean \(\bar{x}\) is:\[\bar{x} = \frac{\text{sum of all data points}}{n}\]For our dataset, the calculated sample mean is \(25.1025\).
  • This mean acts as the best unbiased estimate of the population mean when sampling randomly.
  • It helps us understand the average performance of the sampled students as a group.
The closer the sample mean is to the population mean, the more representative the sample is of the overall population. This is why ensuring random sampling and a sufficiently large sample size is crucial, as it guarantees that the sample mean closely aligns with the actual population characteristics.
Standard Deviation
Standard deviation is a key concept when discussing the variability and spread of a dataset. It quantifies how much individual data points differ from the mean.
A low standard deviation indicates that data points are generally close to the mean, while a high standard deviation shows a wider spread around the mean.
The formula for standard deviation (\(s\)) in a sample is:\[s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}\]Where \(x_i\) stands for each data point, \(\bar{x}\) is the sample mean, and \(n\) is the sample size.
For the ACT scores, the calculated standard deviation is \(2.676\), suggesting moderate variability in scores among the students.
  • Understanding standard deviation is necessary to grasp how spread out the data is, adding context to the mean.
  • It’s pivotal in calculating other statistics like the standard error, used in forming confidence intervals.
Grasping standard deviation enables us to finely interpret data distributions, understanding if observations are tightly clustered or broadly dispersed, deeply impacting statistical conclusions.

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Most popular questions from this chapter

The article "Extravisual Damage Detection? Defining the Standard Normal Tree" (Photogrammetric Engrg. Remote Sensing, 1981: 515-522) discusses the use of color infrared photography in identification of normal trees in Douglas fir stands. Among data reported were summary statistics for green-filter analytic optical densitometric measurements on samples of both healthy and diseased trees. For a sample of 69 healthy trees, the sample mean dye-layer density was \(1.028\), and the sample standard deviation was .163. a. Calculate a \(95 \%\) (two-sided) CI for the true average dye-layer density for all such trees. b. Suppose the investigators had made a rough guess of 16 for the value of \(s\) before collecting data. What sample size would be necessary to obtain an interval width of \(.05\) for a confidence level of \(95 \%\) ?

The article "Evaluating Tunnel Kiln Performance" (Amer. Ceramic Soc. Bull., Aug. 1997: 59-63) gave the following summary information for fracture strengths (MPa) of \(n=169\) ceramic bars fired in a particular kiln: \(x=89.10\), \(s=3.73\). a. Calculate a (two-sided) confidence interval for true average fracture strength using a confidence level of \(95 \%\). Does it appear that true average fracture strength has been precisely estimated? b. Suppose the investigators had believed a priori that the population standard deviation was about \(4 \mathrm{MPa}\). Based on this supposition, how large a sample would have been required to estimate \(\mu\) to within \(.5 \mathrm{MPa}\) with \(95 \%\) confidence?

Even as traditional markets for sweetgum lumber have declined, large section solid timbers traditionally used for construction bridges and mats have become increasingly scarce. The article "Development of Novel Industrial Laminated Planks from Sweetgum Lumber" (J. of Bridge Engr., 2008: 64-66) described the manufacturing and testing of composite beams designed to add value to low- grade sweetgum lumber. Here is data on the modulus of rupture (psi; the article contained summary data expressed in MPa): \(\begin{array}{llllll}6807.99 & 7637.06 & 6663.28 & 6165.03 & 6991.41 & 6992.23 \\ 6981.46 & 7569.75 & 7437.88 & 6872.39 & 7663.18 & 6032.28 \\\ 6906.04 & 6617.17 & 6984.12 & 7093.71 & 7659.50 & 7378.61 \\ 7295.54 & 6702.76 & 7440.17 & 8053.26 & 8284.75 & 7347.95 \\ 7422.69 & 7886.87 & 6316.67 & 7713.65 & 7503.33 & 7674.99\end{array}\) a. Verify the plausibility of assuming a normal population distribution. b. Estimate the true average modulus of rupture in a way that conveys information about precision and reliability. c. Predict the modulus for a single beam in a way that conveys information about precision and reliability. How does the resulting prediction compare to the estimate in (b).

For each of 18 preserved cores from oil-wet carbonate reservoirs, the amount of residual gas saturation after a solvent injection was measured at water flood-out. Observations, in percentage of pore volume, were \(\begin{array}{llllll}23.5 & 31.5 & 34.0 & 46.7 & 45.6 & 32.5 \\ 41.4 & 37.2 & 42.5 & 46.9 & 51.5 & 36.4 \\ 44.5 & 35.7 & 33.5 & 39.3 & 22.0 & 51.2\end{array}\) (See "Relative Permeability Studies of Gas-Water Flow Following Solvent Injection in Carbonate Rocks," Soc. Petrol. Eng. J., 1976: 23-30.) a. Construct a boxplot of this data, and comment on any interesting features. b. Is it plausible that the sample was selected from a normal population distribution? c. Calculate a \(98 \%\) CI for the true average amount of residual gas saturation.

Exercise 63 from Chapter 7 introduced "regression through the origin" to relate a dependent variable \(y\) to an independent variable \(x\). The assumption there was that for any fixed \(x\) value, the dependent variable is a random variable \(Y\) with mean value \(\beta x\) and variance \(\sigma^{2}\) (so that \(Y\) has mean value zero when \(x=0\) ). The data consists of \(n\) independent \(\left(x_{i}, Y_{i}\right)\) pairs, where each \(Y_{i}\) is normally distributed with mean \(\beta x_{i}\) and variance \(\sigma^{2}\). The likelihood is then a product of normal pdf's with different mean values but the same variance. a. Show that the mle of \(\beta\) is \(\beta=\Sigma x_{i} Y_{i} / \Sigma x_{i}^{2}\). b. Verify that the mle of (a) is unbiased. c. Obtain an expression for \(V(\hat{\beta})\) and then for \(\sigma_{\bar{\beta}}\). d. For purposes of obtaining a precise estimate of \(\beta\), is it better to have the \(x_{i}{ }^{\prime}\) s all close to 0 (the origin) or quite far from 0? Explain your reasoning. e. The natural prediction of \(Y_{i}\) is \(\hat{\beta} x_{i}\). Let \(S^{2}=\Sigma\left(Y_{i}-\hat{\beta} x_{i}\right)^{2} /(n-1)\) which is analogous to our earlier sample variance \(S^{2}=\Sigma\left(X_{i}-X\right)^{2} /(n-1)\) for a univariate sample \(X_{1}, \ldots, X_{n}\) (in which case \(X\) is a natural prediction for each \(X_{i}\) ). Then it can be shown that \(T=(\hat{\beta}-\beta) /\left(S / \sqrt{\sum x_{i}^{2}}\right)\) has a \(t\) distribution based on \(n-1\) df. Use this to obtain a CI formula for estimating \(\beta\), and calculate a \(95 \%\) CI using the data from the cited exercise.

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