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The Bureau of Meteorology of the Australian Government provided the mean annual rainfall (in millimeters) in Australia 1983-2002 as follows (http://www.bom.gov.au/ climate/change/rain03.txt): $$ \begin{array}{l} 499.2,555.2,398.8,391.9,453.4,459.8,483.7,417.6,469.2 \\ Check the assumption of normality in the population. Construct a \(95 \%\) confidence interval for the mean annual rainfall. 452.4,499.3,340.6,522.8,469.9,527.2,565.5,584.1,727.3 \\ 558.6,338.6 \end{array} $$

Short Answer

Expert verified
Assuming normality holds, the 95% confidence interval reflects the range for the mean rainfall in Australia from 1983-2002.

Step by step solution

01

Organize and Explore the Data

First, list all the rainfall data provided: \(499.2,555.2,398.8,391.9,453.4,459.8,483.7,417.6,469.2,452.4,499.3,340.6,522.8,469.9,527.2,565.5,584.1,727.3,558.6,338.6\). Start by determining the sample size \(n = 20\). Check for any obviously outlier values, which might affect normality assumptions.
02

Check Normality Assumption

Use a normality test (such as Shapiro-Wilk) or graphical methods like Q-Q plot or histogram to assess if the data is approximately normal. If tests suggest normal distribution, proceed; if not, consider potential data transformation or note the limitation.
03

Calculate Sample Statistics

Compute the sample mean \(\bar{x}\) and sample standard deviation \(s\) of the dataset. For this data set: \[ \bar{x} = \frac{\sum x}{n} \] and \[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \].
04

Determine the Confidence Interval Formula

For a 95% confidence interval, use \(\bar{x} \pm t_{\alpha/2, n-1} \times \frac{s}{\sqrt{n}}\), where \(t_{\alpha/2, n-1}\) is the t-score for \(n-1\) degrees of freedom.
05

Calculate the Confidence Interval

With \(n = 20\), determine \(t_{0.025, 19}\) from a t-distribution table. Calculate the margin of error \( ME = t_{0.025, 19} \times \frac{s}{\sqrt{n}} \). Add and subtract \(ME\) from the sample mean to find the confidence interval: \([\bar{x} - ME, \bar{x} + ME]\).
06

Conclusion

If the confidence interval is computed with a valid normality assumption, interpret the results within the context of the problem, considering variations and stability in annual rainfall.

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Key Concepts

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

Normality Assumption
When analyzing statistical data, one key step is to verify if the data follows a normal distribution. This is what we call checking the "normality assumption."
Normal distribution is essential because many statistical methods, like calculating confidence intervals, rely on this assumption. It provides a symmetric bell-shaped curve when data points are plotted.
To determine if data is normally distributed, we can use visual methods like plotting a histogram or Q-Q plot. These graphs help us see if the data falls along a straight diagonal line, which indicates normality.

Assuming normality allows us to apply parametric tests more effectively. However, if our data significantly deviates from a normal distribution, it might affect the conclusions we draw, necessitating alternative approaches such as data transformation or using non-parametric methods.
Shapiro-Wilk Test
The Shapiro-Wilk test is a statistical test specifically designed to assess the normality of data. It examines the null hypothesis that a sample dataset comes from a normally distributed population. Essentially, it tells us how much our data resembles a normal distribution.

To perform the Shapiro-Wilk test, you compute the test statistic, which is based on the correlations between the data and the corresponding normal scores. If the test result shows a p-value less than a chosen significance level (commonly 0.05), we reject the null hypothesis. This means our data does not follow a normal distribution.

One of the advantages of this test is its sensitivity to small deviations from normality, which makes it a powerful tool for smaller samples. However, it is always beneficial to use it in conjunction with graphical checks to fully understand how your data behaves.
Sample Mean
The sample mean is an important statistic that represents the average of a set of observations. It is calculated by summing up all the data points and then dividing by the number of observations, denoted by:
  • displaying as \(\bar{x} = \frac{\sum x}{n}\),where \(\bar{x}\) is the sample mean, \(\sum x\) is the sum of all data points, and \(n\) is the sample size.
The sample mean serves as an estimation of the population mean when the population is too large to measure completely.Being a measure of central tendency, it provides valuable insight into the data set's overall behavior.The sample mean is central to constructing confidence intervals, which is crucial for estimating population parameters with the associated level of uncertainty.
Sample Standard Deviation
The sample standard deviation is a measure that quantifies the amount of variation or dispersion of a set of data points. It essentially tells how much individual data points deviate from the sample mean. The formula for calculating the sample standard deviation is:
  • Expressed as \( s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \)where \( s \) is the sample standard deviation, \( x_i \) indicates each individual data point, \( \bar{x} \) is the sample mean, and \( n-1 \) is the degrees of freedom.
A large standard deviation indicates a lot of variation in the data points, while a small one suggests that the data points are closely clustered around the mean.Understanding the sample standard deviation is crucial as it helps in determining the margin of error in confidence intervals and allows us to assess the reliability of the mean as a representative value for the data set.

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Most popular questions from this chapter

Consider the one-sided confidence interval expressions for a mean of a normal population. (a) What value of \(z_{\alpha}\) would result in a \(90 \%\) CI? (b) What value of \(z_{\alpha}\) would result in a \(95 \%\) CI? (c) What value of \(z_{\alpha}\) would result in a \(99 \%\) CI?

An article in Knee Surgery, Sports Traumatology, Arthroscopy ["Arthroscopic Meniscal Repair with an Absorbable Screw: Results and Surgical Technique" (2005, Vol. 13, pp. \(273-279)]\) showed that only 25 out of 37 tears \((67.6 \%)\) located between 3 and \(6 \mathrm{~mm}\) from the meniscus rim were healed. (a) Calculate a two-sided \(95 \%\) confidence interval on the proportion of such tears that will heal. (b) Calculate a \(95 \%\) lower confidence bound on the proportion of such tears that will heal.

The U.S. Postal Service (USPS) has used optical character recognition (OCR) since the mid-1960s. In \(1983,\) USPS began deploying the technology to major post offices throughout the country. Suppose that in a random sample of 500 handwritten zip code digits, 466 were read correctly. (a) Construct a \(95 \%\) confidence interval for the true proportion of correct digits that can be automatically read. (b) What sample sample size is needed to reduce the margin of error to \(1 \% ?\) (c) How would the answer to part (b) change if you had to assume that the machine read only one-half of the digits correctly?

An article in the Australian Journal of Agricultural Research ["Non-Starch Polysaccharides and Broiler Performance on Diets Containing Soyabean Meal as the Sole Protein Concentrate" \((1993,\) Vol. \(44(8),\) pp. \(1483-1499)]\) determined that the essential amino acid (Lysine) composition level of soybean meals is as shown here \((\mathrm{g} / \mathrm{kg})\) $$ \begin{array}{|l|l|l|l|l|} \hline 22.2 & 24.7 & 20.9 & 26.0 & 27.0 \\ \hline 24.8 & 26.5 & 23.8 & 25.6 & 23.9 \\ \hline \end{array} $$ (a) Construct a \(99 \%\) two-sided confidence interval for \(\sigma^{2}\). (b) Calculate a \(99 \%\) lower confidence bound for \(\sigma^{2}\). (c) Calculate a \(90 \%\) lower confidence bound for \(\sigma\). (d) Compare the intervals that you have computed.

Determine the \(t\) -percentile that is required to construct each of the following two-sided confidence intervals: (a) Confidence level \(=95 \%,\) degrees of freedom \(=12\) (b) Confidence level \(=95 \%,\) degrees of freedom \(=24\) (c) Confidence level \(=99 \%,\) degrees of freedom \(=13\) (d) Confidence level \(=99.9 \%,\) degrees of freedom \(=15\)

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