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When the population distribution is normal and \(n\) is large, the sample standard deviation \(S\) has approximately a normal distribution with \(E(S) \approx \sigma\) and \(V(S) \approx \sigma^{2} /(2 n)\). We already know that in this case, for any \(n, \bar{X}\) is normal with \(E(\bar{X})=\mu\) and \(V(\bar{X})=\sigma^{2} / n\) a. Assuming that the underlying distribution is normal, what is an approximately unbiased estimator of the 99th percentile \(\theta=\mu+2.33 \sigma\) ? b. When the \(X_{i}\) 's are normal, it can be shown that \(\bar{X}\) and \(S\) are independent rv's (one measures location whereas the other measures spread). Use this to compute \(V(\hat{\theta})\) and \(\sigma_{\hat{\theta}}\) for the estimator \(\hat{\theta}\) of part (a). What is the estimated standard error \(\hat{\sigma}_{\hat{\theta}}\) ? c. Write a test statistic for testing \(H_{0}: \theta=\theta_{0}\) that has approximately a standard normal distribution when \(H_{0}\) is true. If soil \(\mathrm{pH}\) is normally distributed in a certain region and 64 soil samples yield \(\bar{x}=6.33, s=.16\), does this provide strong evidence for concluding that at most \(99 \%\) of all possible samples would have a \(\mathrm{pH}\) of less than 6.75? Test using \(\alpha=.01\).

Step by step solution

01

Identify the Unbiased Estimator

Given that the population distribution is normal and the parameters are approximately unbiased, the 99th percentile is given by \( \theta = \mu + 2.33 \sigma \). To estimate this, we use the sample mean \( \bar{X} \) as an estimate for \( \mu \) and the sample standard deviation \( S \) as an estimate for \( \sigma \). Thus, the estimator \( \hat{\theta} \) for \( \theta \) is \( \hat{\theta} = \bar{X} + 2.33S \). This estimator is approximately unbiased because \( E(\bar{X}) = \mu \) and \( E(S) \approx \sigma \).

02

Compute the Variance of the Estimator

Since \( \bar{X} \) and \( S \) are independent, the variance of \( \hat{\theta} = \bar{X} + 2.33S \) is given by:\[ V(\hat{\theta}) = V(\bar{X}) + (2.33)^2 V(S) \]We know \( V(\bar{X}) = \sigma^2 / n \) and \( V(S) \approx \sigma^2 / (2n) \). Therefore,\[ V(\hat{\theta}) = \frac{\sigma^2}{n} + (2.33)^2 \frac{\sigma^2}{2n} \].

03

Calculate the Standard Error

The standard deviation of \( \hat{\theta} \) is the square root of its variance:\[ \sigma_{\hat{\theta}} = \sqrt{ V(\hat{\theta}) } = \sqrt{ \frac{\sigma^2}{n} + (2.33)^2 \frac{\sigma^2}{2n} } \]

04

Use Given Data to Find the Estimated Standard Error

Given \( \bar{x} = 6.33 \), \( s = 0.16 \), and \( n = 64 \), substitute these into the variance formula to estimate:\[ \hat{V}(\hat{\theta}) = \frac{s^2}{64} + (2.33)^2 \frac{s^2}{2*64} \]Compute \( \hat{\sigma}_{\hat{\theta}} = \sqrt{\hat{V}(\hat{\theta})} \).

05

Formulate the Test Statistic

The test statistic for \( H_0: \theta = \theta_0 \) is given by:\[ Z = \frac{\hat{\theta} - \theta_0}{\hat{\sigma}_{\hat{\theta}}} \]The null hypothesis is tested using standard normal distribution since the test statistic is approximately normal.

06

Conduct the Hypothesis Test

Substitute \( \hat{\theta} = 6.33 + 2.33 * 0.16 \) to find \( \hat{\theta} \), and calculate \( Z \) using \( \theta_0 = 6.75 \). Compare the \( Z \)-value with the critical value for \( \alpha = 0.01 \) (approximately \( Z = 2.33 \) for one-tailed test). If the computed \( Z \) is greater than the critical value, reject \( H_0 \).

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Key Concepts

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

Normal Distribution

In statistics, the normal distribution is fundamental. It's a probability distribution that is symmetric around the mean, depicted as a bell-shaped curve. The normal distribution is significant due to the central limit theorem, which states that the means of random samples from any distribution will approximate a normal distribution as the sample size becomes large. In the context of estimating percentiles, such as the 99th percentile, we leverage this property of normal distribution. This makes it easier to make inferences about the population from which samples are drawn. Often denoted as \ \( \mathcal{N}(\mu, \sigma^2) \ \), a normal distribution has two parameters: the mean \ \( \mu \ \) and variance \ \( \sigma^2 \ \). The mean indicates the central value, while the variance highlights the spread of the data. This characteristic curve allows for various statistical techniques, especially hypothesis testing, because any linear transformation of a normally distributed variable remains normal.

Variance Calculation

Variance is a crucial concept in statistics as it quantifies the dispersion of data points in a data set. In simple terms, variance measures how spread out the numbers in a data set are. Its formula for a sample is: \ \[ s^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2 \ \] where \ \( x_i \ \) are the data points, \ \( \bar{x} \ \) is the sample mean, and \ \( n \ \) is the number of observations. Variance is always non-negative, because every term in the variance sum is squared, so no negative numbers can exist. A high variance indicates that the data points are widely scattered, while a low variance indicates they are closely packed. In our exercise, we compute the variance of the estimator \ \( \hat{\theta} \ \) which depends on the sum of variance of independent variables. This showcases the additive nature of variance when it comes to independent random variables.

Standard Error

The standard error quantifies how much a sample statistic (like the sample mean or sample standard deviation) is expected to vary due to random sampling. It's essentially the standard deviation of its sampling distribution. For the sample mean, the standard error is computed as \ \( SE = \frac{\sigma}{\sqrt{n}} \ \), where \ \( \sigma \ \) is the population standard deviation and \ \( n \ \) is the sample size. In the given problem, a formula involving variance helps us compute the estimated standard error for \ \( \hat{\theta} \ \). This is pivotal to gauge the accuracy of our estimator. The standard error decreases as the sample size increases, indicating more reliable estimates. It's a critical component in constructing confidence intervals and hypothesis tests, providing insights into the reliability of our inferential statistics.

Hypothesis Testing

Hypothesis testing is a statistical method for making decisions using data. Essentially, it allows us to test assumptions (hypotheses) about a population parameter. Here's how it works:

• First, state two hypotheses: the null hypothesis \ \( H_0 \ \) (a baseline assertion) and the alternative hypothesis \ \( H_1 \ \) (where evidence aspires to prove).
• Choose a significance level, \ \( \alpha \ \), often set at 0.05 or 0.01, determining the probability of erroneously rejecting the null hypothesis.
• Calculate a test statistic, typically a Z-score or t-score, which measures how far your sample statistic is from the null hypothesis in units of standard error.
• Compare this calculated statistic to a critical value from a statistical distribution (normal or t-distribution) to decide whether to reject \ \( H_0 \ \).

In our example, we test whether the soil pH is less than a certain value for 99% of samples, comparing the calculated Z-value to a critical value at \ \( \alpha = 0.01 \ \). This powerful method helps scientists and analysts make decisions backed by data.