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Chapter 10: Problem 53

A producer of machine parts claimed that the diameters of the connector rods produced by his plant had a variance of at most .03 inch \(^{2}\). A random sample of 15 connector rods produced by his plant produced a sample mean and variance of .55 inch and .053 inch \(^{2},\) respectively. a. Is there sufficient evidence to reject his claim at the \(\alpha=.05\) level of significance? b. Find a \(95 \%\) confidence interval for the variance of the rod diameters.

Short Answer

Expert verified
What is the 95% confidence interval for the population variance? Answer: Yes, there is sufficient evidence to reject the producer's claim at the 伪=0.05 level of significance. The 95% confidence interval for the population variance is (0.03097, 0.12185).

Step by step solution

01

State the hypotheses

The null hypothesis: The population variance, 蟽^2, is at most 0.03 inch^2 (蟽^2 鈮� 0.03) The alternative hypothesis: The population variance, 蟽^2, is greater than 0.03 inch^2 (蟽^2 > 0.03)
02

Apply the Chi-square test statistic

To calculate the test statistic, we will use the sample size (n = 15), sample variance (s^2 = .053) and the population variance according to the null hypothesis (蟽^2 = 0.03). The test statistic is calculated as: Chi-square test statistic, \(蠂^2 = (n-1) * (s^2 / 蟽^2) = (15-1) * (0.053 / 0.03)\)
03

Calculate the critical value

We need to find the critical value for a chi-square distribution with 14 degrees of freedom and a significance level of 伪 = 0.05. The rejection region for the null hypothesis is in the upper tail, so we will look up the value for \(蠂^2_{1 - 伪, df}\): \(蠂^2_{0.95,14}\)
04

Compare the test statistic to the critical value

If the calculated chi-square test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
05

Find a 95% confidence interval for the variance

To find a 95% confidence interval for the population variance, we will use the sample variance (s^2 = 0.053) and the chi-square distribution with 14 degrees of freedom. The confidence interval is given by: \(((n-1)s^2) / (蠂^2_{1 - 伪/2}) 鈮� 蟽^2 鈮� ((n-1)s^2) / (蠂^2_{伪/2})\) Substitute the given values and calculate the confidence interval. #Solution#
06

Calculate the Chi-square test statistic

Chi-square test statistic, \(蠂^2 = (15-1) * (0.053 / 0.03) = 14 * (0.053 / 0.03) \approx 24.53\)
07

Find the critical value

From the chi-square distribution table, \(蠂^2_{0.95,14} \approx 23.685\)
08

Compare the test statistic to the critical value

Since the calculated test statistic (24.53) is greater than the critical value (23.685), we reject the null hypothesis. Therefore, there is sufficient evidence to reject the producer's claim at the 伪=0.05 level of significance.
09

Calculate the 95% confidence interval for the variance

Using the chi-square distribution table, we find \(蠂^2_{0.025,14} \approx 6.571\) and \(蠂^2_{0.975,14} \approx 29.141\) \(((15-1)*0.053) / 29.141 鈮� 蟽^2 鈮� ((15-1)*0.053) / 6.571\) \(0.03097 鈮� 蟽^2 鈮� 0.12185\) The 95% confidence interval for the population variance is (0.03097, 0.12185).

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a fundamental part of statistical analysis, used to make informed conclusions about a population parameter based on sample data. In any hypothesis test, two competing statements or hypotheses are considered:
  • The null hypothesis (\(H_0\)) represents the status quo or a statement of no effect or difference. In the example, \(H_0\) claims the variance is at most 0.03 inch虏.
  • The alternative hypothesis (\(H_a\)) reflects a new claim or the opposite of the null. Here, \(H_a\) suggests the variance is greater than 0.03 inch虏.
By comparing observed data with what we would expect under the null, we decide whether to reject the null hypothesis, often using a test statistic, and a predefined level of significance.
Chi-square Distribution
The chi-square distribution is heavily used in hypothesis testing, especially when dealing with variances. It is a continuous probability distribution that is particularly suited for tests of variance. In the exercise, the chi-square test statistic was calculated using the formula:\[蠂^2 = (n-1) \times \frac{s^2}{蟽^2}\]where:
  • \(n\) is the sample size.
  • \(s^2\) represents the sample variance.
  • \(蟽^2\) is the hypothesized population variance.
The resulting chi-square statistic helps us compare whether the sample variance significantly differs from the hypothesized variance, guiding us on rejecting or failing to reject the null hypothesis.
Confidence Interval
A confidence interval provides a range of plausible values for a population parameter, offering more insight than a single estimate alone. To calculate a confidence interval for variance using the chi-square distribution, we employ the formula:\[((n-1)s^2) / 蠂^2_{1 - 伪/2} \leq 蟽^2 \leq ((n-1)s^2) / 蠂^2_{伪/2}\]This approach yields a range in which the true population variance likely falls. In this case study, a 95% confidence interval means that if we were to draw many samples and calculate a confidence interval from each, about 95% of those intervals would contain the true variance.
Variance
Variance measures how much values in a data set differ from the mean. It provides an indication of data spread or dispersion. A higher variance means data are more spread out, while lower variance indicates data points are closer to the mean. In the context of this exercise, the producer claims the variance of the diameters is at most 0.03 inch虏, indicating a belief in a controlled, consistent production process. However, the sample variance obtained was 0.053 inch虏, suggesting more variability than the claimed specification.
Significance Level
The significance level, denoted as \(伪\), is the threshold used to decide whether to reject the null hypothesis. Commonly set at 0.05, it represents a 5% risk level of rejecting the null hypothesis when it is true (Type I error). In this exercise, a 0.05 significance level was used, meaning there must be strong enough evidence against the null for it to be rejected. When the test statistic exceeds the critical value from the chi-square distribution table corresponding to this significance level, it leads us to reject the null hypothesis, showing that the sample results are unlikely to have occurred under the assumption of \(H_0\).

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Most popular questions from this chapter

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