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Chapter 11: Problem 47

The makers of Flippin' Out Pancake Mix claim that one cup of their mix contains 11 grams of sugar. However, the mix is not uniform, so the amount of sugar varies from cup to cup. One cup of mix was taken from each of 24 randomly selected boxes. The sample variance of the sugar measurements from these 24 cups was \(1.47\) grams. Assume that the distribution of sugar content is approximately normal. a. Construct the \(98 \%\) confidence intervals for the population variance and standard deviation. b. Test at the \(1 \%\) significance level whether the variance of the sugar content per cup is greater than \(1.0\) gram.

Short Answer

Expert verified
a. The 98% confidence interval for the population variance is (0.907, 3.07) and for the standard deviation is (0.952, 1.752). b. There is not sufficient evidence to conclude that the variance of the sugar content per cup is greater than 1 gram at the 1% significance level.

Step by step solution

01

Understanding Variables

The sample size (\(n\)) given is 24, and the sample variance (\(s^{2}\)) is 1.47.
02

Construct Confidence Interval for Variance

To construct a \(98 \%\) confidence interval, we need to find the chi-square critical values. These are obtained from the chi-square distribution table. The degrees of freedom equal to \(df = n-1 = 24-1 = 23\). For a two-sided confidence interval, we split the \(伪 = 1 - 0.98 = 0.02\) into two, meaning \(伪/2 = 0.01\). Thus, the critical values \(\chi^{2}_{(a/2, df)} = 11.689\) and \(\chi^{2}_{(1-a/2, df)} = 38.076\). Plug in the obtained chi-square values and the test statistic into the formula for the confidence interval for variance: \((n-1)s^{2}/\chi^{2}_{(1-a/2, df)}, (n-1)s^{2}/\chi^{2}_{(a/2, df)}\), which gives us (0.907, 3.07)
03

Construct Confidence Interval for Standard Deviation

The confidence interval for standard deviation is derived from the confidence interval for variance. It's the square root of the variance interval endpoints. This gives us the interval: (\(0.952\), \(1.752\)).
04

Test Hypothesis for Variance

We want to test whether the variance is greater than \(1.0\) gram. Our null hypothesis (\(H_{0}\)) is that the variance is \(1.0\) and the alternative hypothesis (\(H_{1}\)) is that the variance is greater than \(1.0\). We use a Chi-square test to compute the test statistic \(\chi^{2} = (n-1)s^{2}/\sigma_{0}^{2}\), where \(\sigma_{0}^{2}\) is the assumed population variance under \(H_{0}\). This gives \(\chi^{2} = 31.776\). We compare this with the critical value \(\chi^{2}_{(1-\alpha, df)} = 36.415\). Since \(\chi^{2} < \chi^{2}_{(1-\alpha, df)}\), we fail to reject the null hypothesis and conclude that there is not sufficient evidence to say that the variance is greater than \(1\) at the \(1 \%\) significance level.

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Key Concepts

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

Confidence Intervals
A confidence interval provides a range in which we expect a population parameter to lie with a certain level of confidence. In this case, we're looking at the population variance and standard deviation. We've constructed a 98% confidence interval for both:
For the variance, using the chi-square distribution, we calculated the interval as
  • Lower bound: 0.907
  • Upper bound: 3.07
For the standard deviation, we take the square root of these values:
  • Lower bound: 0.952
  • Upper bound: 1.752
These intervals tell us that we are 98% confident that the true population variance and standard deviation lie within these ranges.
Chi-square Distribution
The chi-square distribution is vital when dealing with variance in statistics. It's a skewed distribution that becomes more symmetrical with larger sample sizes. For our exercise, with 23 degrees of freedom, we use the chi-square table to find critical values.
Chi-square values depend on the confidence level and sample size. Here, for a 98% confidence interval, we found critical values:

  • Lower critical value: 11.689
  • Upper critical value: 38.076
These values are then used in formulas to construct confidence intervals for variance, supporting our ability to make inferences about the population.
Hypothesis Testing
Hypothesis testing allows us to make informed decisions based on sample data. In our problem, we tested whether the variance of sugar content is more than 1 gram.
Steps in hypothesis testing:
  • Set up null (



    H_0 ) and alternative (



  • H_1) hypotheses: \(H_0: \sigma^2 = 1\) and \(H_1: \sigma^2 > 1\).
  • Calculate the test statistic using the formula \(\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}\) which gives us 31.776.
  • Compare the test statistic with critical chi-square value of 36.415 at \(1\%\) significance level.
  • Decision rule: If \(\chi^2\) is less than the critical value, we fail to reject \(H_0\).
Since our test statistic is less than the critical value, we conclude there's not enough evidence to say the variance exceeds 1 gram.
Sample Variance
Sample variance \(s^2\) measures how much the data values in a sample deviate from their mean. It's a crucial tool for understanding data spread, especially when estimating the population variance.
In our scenario, the sample variance of sugar content was 1.47 grams. This value helps us in several ways:
  • Used to construct confidence intervals, giving us an interval estimate of the population variance.
  • In hypothesis testing, contributing to the calculation of the test statistic, which determines how extreme our sample variance is compared to a hypothesized population variance.
The concept of sample variance is foundational in statistics, providing insights into data variability and underpinning many statistical procedures.

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Most popular questions from this chapter

Determine the value of \(\chi^{2}\) for 23 degrees of freedom and an area of \(.990\) in the left tail of the chisquare distribution curve.

Construct the \(95 \%\) confidence intervals for the population variance and standard deviation for the following data, assuming that the respective populations are (approximately) normally distributed. a. \(n=10, s^{2}=7.2\) b. \(n=18, s^{2}=14.8\)

A sample of 21 observations selected from a normally distributed population produced a sample variance of \(1.97\). a. Write the null and alternative hypotheses to test whether the population variance is greater than \(1.75\). b. Using \(\alpha=.025\), find the critical value of \(\chi^{2}\). Show the rejection and nonrejection regions on a chi-square distribution curve. c. Find the value of the test statistic \(x^{2}\). d. Using the \(2.5 \%\) significance level, will you reject the null hypothesis stated in part a?

Suppose that you have a two-way table with the following row and column totals. \begin{tabular}{l|cccc|c} \hline & & \multicolumn{3}{c|} { Variable 1 } & \\ \cline { 3 - 5 } \multicolumn{2}{l|} {} & A & B & C & Total \\ \hline & \(\mathbf{X}\) & & & & 120 \\ Variable 2 & \(\mathbf{Y}\) & & & & 205 \\ & \(\mathbf{Z}\) & & & & 175 \\ \hline & Total & 165 & 140 & 195 & 500 \\ \hline \end{tabular} The observed values in the cells must be counts, which are nonnegative integers. Calculate the expected counts for the cells under the assumption that the two variables are independent. Based on your calculations, explain why it is impossible for the test statistic to have a value of zero.

The variance of the SAT scores for all students who took that test this year is 5000 . The variance of the SAT scores for a random sample of 20 students from one school is equal to 3175 . a. Test at the \(2.5 \%\) significance level whether the variance of the SAT scores for students from this school is lower than 5000 . Assume that the SAT scores for all students at this school are (approximately) normally distributed. b. Construct the \(98 \%\) confidence intervals for the variance and the standard deviation of SAT scores for all students at this school.
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