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Instrument Precision A precision instrument is guaranteed to read accurately to within 2 units. A sample of four instrument readings on the same object yielded the measurements \(353,351,351,\) and \(355 .\) a. Test the null hypothesis that \(\sigma=.7\) against the alternative \(\sigma>.7\). Use \(\alpha=.05 .\) b. Find a \(90 \%\) confidence interval for the population variance.

Short Answer

Expert verified
Based on the given instrument readings, we conducted a hypothesis test that rejected the null hypothesis, providing sufficient evidence to support the alternative hypothesis that the population standard deviation is greater than $0.7$. Furthermore, we calculated a $90\%$ confidence interval for the population variance to be approximately $(1.54, 20.69)$.

Step by step solution

01

Calculate the sample mean and variance

Firstly, find the sample mean (\(\bar{X}\)) and sample variance (\(s^2\)) using the given readings: \(\bar{X} = \frac{353+351+351+355}{4} = 352.5\) \(s^2 = \frac{\sum_{i=1}^{4}(X_i - \bar{X})^2}{4-1} = \frac{(353-352.5)^2+(351-352.5)^2+(351-352.5)^2+(355-352.5)^2}{3} = 4\)
02

Find the chi-square test statistic

Next, calculate the chi-square test statistic (\(\chi_0^2\)) using the sample variance and null hypothesis standard deviation: \(\chi_0^2 = \frac{(n-1) \cdot s^2}{\sigma_0^2} = \frac{(4-1) \cdot 4}{0.7^2} \approx 16.33\)
03

Determine the rejection region

We are given a significance level of \(\alpha = 0.05\). Since we are testing the null hypothesis against a one-sided alternative hypothesis, we need to find the critical value \(\chi^2_{\alpha,v}\) from the Chi-square distribution: \(\chi^2_{0.05, 3} \approx 7.81\) (from the Chi-square distribution table) We will reject the null hypothesis if the test statistic \(\chi_0^2\) is greater than the critical value \(\chi^2_{0.05, 3}\).
04

Perform the hypothesis test

Since the test statistic is \(\chi_0^2 \approx 16.33\), which is greater than the critical value \(\chi^2_{0.05, 3} \approx 7.81\), we reject the null hypothesis. There is enough evidence to support the alternative hypothesis that \(\sigma > 0.7\).
05

Find the confidence interval for population variance

We are asked to find a \(90\%\) confidence interval for the population variance. To do so, we need to find the values \(\chi^2_{0.05,3}\) and \(\chi^2_{0.95,3}\) from the Chi-square distribution table: \(\chi^2_{0.05, 3} \approx 7.81\) \(\chi^2_{0.95, 3} \approx 0.58\) Now, the \(90\%\) confidence interval can be calculated as follows: \(\left( \frac{(n-1)s^2}{\chi^2_{0.05, 3}}, \frac{(n-1)s^2}{\chi^2_{0.95, 3}} \right) = \left( \frac{3\cdot4}{7.81}, \frac{3\cdot4}{0.58} \right) \approx (1.54, 20.69)\) Therefore, the \(90\%\) confidence interval for the population variance is \((1.54, 20.69)\).

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Key Concepts

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

Chi-Square Test
The Chi-Square Test is a statistical method used to determine if there is a significant difference between expected and observed frequencies in one or more categories. In the context of hypothesis testing, it can also be used to assess the variance of a population. When you perform a Chi-Square Test to compare the sample variance to a hypothesized population variance, you are essentially checking if the variability you observe in your data is likely given your assumption about the population variance.

For example, let's say you're given a hypothesis that an instrument has a population standard deviation \(\sigma = 0.7\). By measuring the variance from a sample and calculating the Chi-Square Test statistic, you compare this statistic to a critical value from the Chi-square distribution. If the test statistic exceeds the critical value, the null hypothesis that the population variance is equal to the hypothesized variance is rejected. This is exactly what happened in the textbook exercise, where the calculated Chi-Square Test statistic was approximately 16.33, higher than the critical value of 7.81 for \(\alpha = 0.05\) and 3 degrees of freedom, leading to the conclusion that the true population variance is likely greater than \(0.7^2\).
Confidence Interval
A Confidence Interval (CI) is a range of values that is likely to contain the population parameter with a specified level of confidence. It provides an estimated range constructed from sample data and gives us an idea about where the true population parameter lies. Confidence levels most commonly used are 90%, 95%, and 99%. The higher the confidence level, the wider the confidence interval.

In the textbook exercise, the task was to find a 90% confidence interval for the population variance. Using the Chi-Square distribution and the sample variance calculated from the data, the confidence interval was determined by finding Chi-Square values corresponding to the lower and upper ends of the 90% confidence level. These values are then used to calculate the lower and upper bounds of the interval for the population variance, leading to a range of approximately \(1.54, 20.69\). This interval indicates that we can be 90% confident that the actual variance of the population falls within this range.
Population Variance
Population Variance is a measure of the spread between numbers in a data set. It represents the average of the squared differences from the mean, providing an overall measure of the variability within a dataset. Variance is a powerful concept in statistics as it gives a clear picture of data distribution and is the foundation of many statistical tests, including the Chi-Square Test.

In the given exercise, the population variance is what we are trying to determine or estimate. The sample variance is calculated from measurable data, and then from that, we infer the population variance. Calculating the sample variance involves finding the mean of the sample, subtracting the sample mean from each sample observation, squaring each result, summing them up, and dividing by the number of observations minus one. Finally, the textbook exercise required testing if the computed sample variance significantly differed from the assumed population variance (\(\sigma^2 = 0.7^2\)) using a chi-square test. The resulting analysis gives insights into the precision of the instrument being tested.

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Most popular questions from this chapter

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In response to a complaint that a particular tax assessor (1) was biased, an experiment was conducted to compare the assessor named in the complaint with another tax assessor (2) from the same office. Eight properties were selected, and each was assessed by both assessors. The assessments (in thousands of dollars) are shown in the table. $$\begin{array}{ccc}\hline \text { Property } & \text { Assessor } 1 & \text { Assessor } 2 \\\\\hline 1 & 276.3 & 275.1 \\\2 & 288.4 & 286.8 \\\3 & 280.2 & 277.3 \\ 4 & 294.7 & 290.6 \\\5 & 268.7 & 269.1 \\\6 & 282.8 & 281.0 \\\7 & 276.1 & 275.3 \\\8 & 279.0 & 279.1 \\\\\hline\end{array}$$ a. Do the data provide sufficient evidence to indicate that assessor 1 tends to give higher assessments than assessor 2 ? b. Estimate the difference in mean assessments for the two assessors. c. What assumptions must you make in order for the inferences in parts a and b to be valid?

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