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A random sample of size \(n=7\) from a normal population produced these measurements: \(1.4,3.6,1.7,\) \(2.0,3.3,2.8,2.9 .\) a. Calculate the sample variance, \(s^{2}\) b. Construct a \(95 \%\) confidence interval for the population variance, \(\sigma^{2}\). c. Test \(H_{0}: \sigma^{2}=.8\) versus \(H_{\mathrm{a}}: \sigma^{2} \neq .8\) using \(\alpha=.05 .\) State your conclusions. d. What is the approximate \(p\) -value for the test in part \(\mathrm{c} ?\)

Short Answer

Expert verified
Based on the given random sample, the sample variance is approximately 0.77, and the 95% confidence interval for the population variance is (0.32, 3.72). Using a significance level of 0.05, we fail to reject the null hypothesis that the population variance equals 0.8. There is not enough evidence to support the alternative hypothesis that the population variance is different from 0.8. The approximate p-value for this test is 0.33.

Step by step solution

01

Find the sample mean and sample variance

Calculate the sample mean (\(\bar{x}\)) and sample variance (\(s^2\)) using the given sample data: Sample data: \(1.4, 3.6, 1.7, 2.0, 3.3, 2.8, 2.9\) Sample mean: \(\bar{x} = \frac{1.4+3.6+1.7+2.0+3.3+2.8+2.9}{7} = \frac{17.7}{7} = 2.53\) Sample variance: \(s^2 = \frac{\Sigma(x_i - \bar{x})^2}{n-1} = \frac{(1.4-2.53)^2+(3.6-2.53)^2+\cdots+(2.9-2.53)^2}{7-1} = \frac{4.61}{6} = 0.77\)
02

Construct a 95% confidence interval for the population variance

To construct a 95% confidence interval for the population variance (\(\sigma^2\)), we will use the chi-square distribution. The formula for the confidence interval is: \(\left( \frac{(n-1) s^2}{\chi^2_{\alpha/2, n-1}}, \frac{(n-1) s^2}{\chi^2_{1-\alpha/2, n-1}} \right)\) In this case, \(n=7, \alpha=.05,\) and \(s^2=0.77\). Using a chi-square table or calculator, we find the critical values: \(\chi^2_{.025,6} = 1.24\) and \(\chi^2_{.975,6} = 14.45\) Substituting the values, we get the confidence interval: \((\frac{(6)(0.77)}{14.45},\frac{(6)(0.77)}{1.24}) \approx (0.32, 3.72)\)
03

Perform the hypothesis test

We are given the null hypothesis \(H_0: \sigma^2=.8\) and the alternative hypothesis \(H_a: \sigma^2 \neq .8\). To test these hypotheses, we will use the chi-square test statistic: \(\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}\) Where the hypothesized value of the population variance is \(\sigma_0^2 =0.8\). We find the test statistic: \(\chi^2 =\frac{(6) (0.77)}{0.8} \approx 5.78\) Using a chi-square table or calculator, find the corresponding p-value for the test statistic: \(p = P(\chi^2 > 5.78 \,\text{or} \, \chi^2 < 5.78) \approx 0.33\)
04

Make conclusions

The p-value (\(0.33\)) is greater than the significance level (\(0.05\)). Therefore, we fail to reject the null hypothesis \(H_0: \sigma^2 = 0.8\). We can conclude that there is not enough evidence to support the alternative hypothesis \(H_a: \sigma^2 \neq 0.8\).
05

Answer

a. The sample variance is \(s^2 = 0.77\). b. The 95% confidence interval for the population variance is \((0.32, 3.72)\). c. We fail to reject the null hypothesis with a p-value of \(0.33\). There is not enough evidence to support the alternative hypothesis. d. The approximate p-value for the test in part c is \(0.33\).

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Key Concepts

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

Sample Variance

Understanding sample variance is crucial for analyzing data sets and drawing conclusions about the variability within a population. It measures how much the individual data points differ from the sample mean. Mathematically, it is calculated by taking the sum of the squared differences between each data point and the sample mean, and then dividing by one less than the number of data points (this is known as the degrees of freedom). The formula is given by:

\[ s^{2} = \frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^{2}\]

In context, if the sample variance is small, it indicates that the data points are close to the mean, showing low variability, while a large sample variance signifies that the data points are spread out over a wider range of values, indicating high variability.


Relevance in Statistical Analysis

Sample variance is not only important in calculating other statistics like the standard deviation but also lays the foundation for more complex procedures like hypothesis testing. In our exercise, the sample variance was a stepping stone to creating a confidence interval and conducting a hypothesis test about the population variance.

Hypothesis Testing

In the realm of statistics, hypothesis testing is a method for evaluating two mutually exclusive statements about a population to determine which statement is best supported by the sample data. When performing a hypothesis test, we start by stating a null hypothesis (\(H_0\)), which represents a default position or status quo, and an alternative hypothesis (\(H_a\)), which represents what we aim to prove.


Steps in Hypothesis Testing

  • Select the appropriate test based on the data and hypotheses.
  • Determine the level of significance (\(alpha\)), commonly set at 0.05.
  • Calculate the test statistic from the sample data.
  • Compare the test statistic to the critical value(s) or use the corresponding p-value to make a decision.
  • Draw a conclusion based on the comparison.

In our textbook problem, a hypothesis test was conducted for the population variance, where the null hypothesis (\(H_0: \sigma^{2}=.8\)) was tested against an alternative hypothesis that the variance is not equal to 0.8. The test resulted in failing to reject the null hypothesis, implying insufficient evidence against it from the sample provided.

Chi-Square Distribution

The chi-square distribution is a fundamental probability distribution in statistics that arises in tests of independence and hypothesis tests for variance, among other applications. It is skewed right and only defined for positive values. The shape of the chi-square distribution depends on its degrees of freedom (df), with more degrees of freedom leading to a shape closer to a normal distribution.


Usage in Variance Estimation

One critical use of the chi-square distribution is to create confidence intervals for population variance. This application relies on the fact that certain combinations of sample variance and sample size follow a chi-square distribution, provided the sample comes from a normally distributed population. In our exercise, this distribution was used to construct a 95% confidence interval for the population variance, guiding us on the range within which the true variance is likely to fall with high certainty.

P-Value

The p-value is a probability that provides a measure of the evidence against the null hypothesis provided by the sample. It answers the question: 'If the null hypothesis were true, what is the probability that we would observe a test statistic as extreme as the one calculated from our sample data, or more extreme?' A small p-value (typically less than 0.05) indicates strong evidence against the null hypothesis, hence we might reject it favoring the alternative hypothesis.


Interpreting P-Values

In hypothesis testing, a p-value that exceeds the significance level (\(alpha\)) implies that we fail to reject the null hypothesis, as there is not enough evidence against it. Conversely, if the p-value is below \(alpha\), it suggests that the null hypothesis is unlikely to be true. In the given problem, the p-value was approximately 0.33, failing to provide sufficient evidence to reject the null hypothesis, which claimed that the population variance is 0.8.

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Most popular questions from this chapter

Although there are many treatments for bulimia nervosa, some subjects fail to benefit from treatment. In a study to determine which factors predict who will benefit from treatment, an article in the British Journal of Clinical Psychology indicates that self-esteem was one of these important predictors. The table gives the mean and standard deviation of self-esteem scores prior to treatment, at post treatment, and during a follow-up: $$ \begin{array}{lccc} & \text { Pretreatment } & \text { Posttreatment } & \text { Follow-up } \\ \hline \text { Sample Mean } \bar{x} & 20.3 & 26.6 & 27.7 \\ \text { Standard Deviation } s & 5.0 & 7.4 & 8.2 \\ \text { Sample Size } n & 21 & 21 & 20 \end{array} $$ a. Use a test of hypothesis to determine whether there is sufficient evidence to conclude that the true pretreatment mean is less than 25 . b. Construct a \(95 \%\) confidence interval for the true posttreatment mean. c. In Section 10.4 , we will introduce small-sample techniques for making inferences about the difference between two population means. Without the formality of a statistical test, what are you willing to conclude about the differences among the three sampled population means represented by the results in the table?

To properly treat patients, drugs prescribed by physicians must have a potency that is accurately defined. Consequently, not only must the distribution of potency values for shipments of a drug have a mean value as specified on the drug's container, but also the variation in potency must be small. Otherwise, pharmacists would be distributing drug prescriptions that could be harmfully potent or have a low potency and be ineffective. A drug manufacturer claims that his drug is marketed with a potency of \(5 \pm .1\) milligram per cubic centimeter \((\mathrm{mg} / \mathrm{cc}), \mathrm{A}\) random sample of four containers gave potency readings equal to \(4.94,5.09,5.03,\) and \(4.90 \mathrm{mg} / \mathrm{cc}\) a. Do the data present sufficient evidence to indicate that the mean potency differs from \(5 \mathrm{mg} / \mathrm{cc} ?\) b. Do the data present sufficient evidence to indicate that the variation in potency differs from the error limits specified by the manufacturer? [HINT: It is sometimes difficult to determine exactly what is meant by limits on potency as specified by a manufacturer. Since he implies that the potency values will fall into the interval \(5 \pm .1 \mathrm{mg} / \mathrm{cc}\) with very high probability - the implication is always-let us assume that the range \(.2 ;\) or \((4.9\) to 5.1\(),\) represents \(6 \sigma,\) as suggested by the Empirical Rule. Note that letting the range equal \(6 \sigma\) rather than \(4 \sigma\) places a stringent interpretation on the manufacturer's claim. We want the potency to fall into the interval \(5 \pm .1\) with very high probability.

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Calculate \(s^{2}\), the pooled estimator for \(\sigma^{2}\), in these cases: a. \(n_{1}=10, n_{2}=4, s_{1}^{2}=3.4, s_{2}^{2}=4.9\) b. \(n_{1}=12, n_{2}=21, s_{1}^{2}=18, s_{2}^{2}=23\)

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