Question

A chemical manufacturer claims that the purity of his product never varies by more than \(2 \% .\) Five batches were tested and given purity readings of \(98.2,97.1,98.9,97.7,\) and \(97.9 \%\) a. Do the data provide sufficient evidence to contradict the manufacturer's claim? (HINT: To be generous, let a range of \(2 \%\) equal \(4 \sigma .)\) b. Find a \(90 \%\) confidence interval for \(\sigma^{2}\).

Short answer

Answer: Yes, the data provides sufficient evidence to contradict the manufacturer's claim. Question: What is the 90% confidence interval for the variance of the purity readings? Answer: The 90% confidence interval for the variance is (0.3051, 2.7164) or in terms of percentages (0.3051%, 2.7164%).

Step-by-step solution

State the hypotheses

The null hypothesis, \(H_0\), is that the manufacturer's claim is true. The alternative hypothesis, \(H_1\), is that the claim is false. In terms of standard deviation (\(\sigma\)), the hypotheses can be stated as: \(H_0: \sigma = 0.5 \%\) (where \(4 \sigma = 2 \%\) according to the hint) \(H_1: \sigma > 0.5 \%\)

Calculate sample mean and sample variance

Let's calculate the sample mean (\(\bar{x}\)) and sample variance (\(S^2\)) using the purity readings. \(\bar{x} = \frac{1}{5}(98.2 + 97.1 + 98.9 + 97.7 + 97.9) = 97.96 \%\) \(S^2 = \frac{\sum(x_i - \bar{x})^2}{n - 1} = \frac{(98.2-97.96)^2+(97.1-97.96)^2+(98.9-97.96)^2+(97.7-97.96)^2+(97.9-97.96)^2}{5 - 1} = 0.7225\)

Calculate the test statistic

The test statistic \(X^2\) is given by: \(X^2 = \frac{(n-1)S^2}{\sigma^2} = \frac{(5-1)0.7225}{(0.5)^2} = 11.56\)

Find the critical value

For a one-tailed test with 4 degrees of freedom (as \(n - 1 = 4\)) and a significance level of 0.05, find the critical value using chi-square distribution table: \(\chi_{0.05,4}^2 = 9.488\)

Make the decision

As the test statistic \(X^2 = 11.56\) is greater than the critical value \(\chi_{0.05,4}^2 = 9.488\), we reject the null hypothesis. Answer: The data provides sufficient evidence to contradict the manufacturer's claim. #b. Confidence interval#

Find the chi-square values

To find a 90% confidence interval for the variance (\(\sigma^2\)), we need to find the chi-square values corresponding to the lower and upper tail probabilities of 0.05: \(\chi_{0.95,4}^2 = 1.064\) \(\chi_{0.05,4}^2 = 9.488\)

Calculate the confidence interval

Now, use these chi-square values to find the confidence interval for the variance: Lower limit for the variance: \(\frac{(n - 1)S^2}{\chi_{0.05,4}^2} = \frac{(5 - 1)0.7225}{9.488} = 0.3051\) Upper limit for the variance: \(\frac{(n - 1)S^2}{\chi_{0.95,4}^2} = \frac{(5 - 1)0.7225}{1.064} = 2.7164\) Answer: The 90% confidence interval for the variance is \((0.3051, 2.7164)\) or in terms of percentages \((0.3051\%, 2.7164\%)\).

Key concepts

Chi-Square Distribution

The chi-square distribution is an important concept in hypothesis testing, particularly when dealing with variance. It is a tool used to assess how a sample's variance compares to the variance claimed by a population. In a chi-square test, the null hypothesis can suggest that the variance of a dataset is equal to a specific value, and the test can tell if this hypothesis is reasonable.

• The test statistic is calculated as: \[X^2 = \frac{(n-1)S^2}{\sigma^2}\] where \( n \) is the sample size, \( S^2 \) is the sample variance, and \( \sigma^2 \) is the population variance posited by the null hypothesis.
• The degrees of freedom in this test are \( n - 1 \), reflecting the number of values in the final calculation of a statistic that are free to vary.
• Critical values of the chi-square distribution can be found from tables for specific confidence levels and degrees of freedom. These help determine whether the test statistic falls within the region to reject the null hypothesis.

In our case, the chi-square statistic calculated was 11.56, and the critical value for 4 degrees of freedom at a 0.05 significance level was 9.488. Since 11.56 is greater than 9.488, we reject the null hypothesis, suggesting that the variance exceeds the manufacturer's claim.

Confidence Interval

A confidence interval is an estimated range of values which is likely to include an unknown population parameter. When calculating a confidence interval for variance, chi-square values are again crucial.

• To determine the confidence interval for the variance \( \sigma^2 \), one uses chi-square distribution to find two critical values: one for the lower bound and one for the upper bound corresponding to a given confidence level.
• The confidence interval is given by:\[\left( \frac{(n-1)S^2}{\chi_{upper}}, \frac{(n-1)S^2}{\chi_{lower}} \right)\] where \( \chi_{upper} \) and \( \chi_{lower} \) are the chi-square values for the upper and lower tails of the distribution for the confidence level desired.

In this problem, we calculated a 90% confidence interval for the variance of the chemical batch purities. The interval was determined to be between 0.3051 and 2.7164. This range gives us an understanding of the possible variation in purity, providing a statistical measure of the uncertainty or reliability of the variance estimate.

Standard Deviation

Standard deviation is a measure that is used to quantify the amount of variation or dispersion within a set of data values. It is a crucial statistic in understanding how much individual data points in a sample differ from the sample mean.

• In the context of hypothesis testing, especially with regards to variance, standard deviation provides a basis for comparing the sample's dispersion with what is hypothesized about the population.
• The formula for standard deviation \( \sigma \) in a sample is: \[\sigma = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n - 1}}\] where \( x_i \) represents each value in the sample and \( \bar{x} \) is the sample mean.
• When validating claims like our exercise, the evidence comes from seeing if calculated and hypothesized standard deviations match. If they don't, it can indicate that the actual variability differs from what was expected.

In our situation, we tested if the actual standard deviation of the chemical purity exceeded 0.5%, the point where it would contravene the manufacturer's claim of purity variation not exceeding 2%. The hypothesis testing showed that the standard deviation did indeed exceed what was claimed, leading to a conclusion that the claim could be contradicted based on the provided samples.