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To test the hypothesis that the standard deviation on a standard test is \(12,\) a sample of 40 randomly selected students' exams was tested. The sample variance was found to be \(155 .\) Does this sample provide sufficient evidence to show that the standard deviation differs from 12 at the 0.05 level of significance?

Short Answer

Expert verified
No, the sample does not provide sufficient evidence to show that the standard deviation differs from 12 at the 0.05 level of significance.

Step by step solution

01

Identifying the Hypotheses

We're testing a hypothesis about the population's variance and standard deviation, so we need to set up the null hypothesis and the alternative hypothesis. The null hypothesis \(H_{0}:\sigma^{2}=144\) states that the population's variance is \(144 (12^2)\). The alternative hypothesis \(H_{1}:\sigma^{2}\neq 144\) contends that the variance is not 144.
02

Calculating the Test Statistic

The test statistic \(X^2\) for a sample variance is calculated by \((n - 1) * s^2 / \sigma^2\), where \(n\) is the sample size, \(s^2\) is the sample variance, and \(\sigma^2\) is the variance under the null hypothesis. Substituting into the formula, we have \((40-1) * 155 / 144 = 40.069.\
03

Identifying the Critical Values

The test statistic follows a chi-square distribution with \(n-1\) degrees of freedom, which is \(40-1=39\) in this case. The critical values at a 0.05 significance level (2-tailed test) are the values at which the lower 0.025 area of the distribution and the upper 0.025 area. Using a chi-square distribution table, the critical values for \(df=39\) are \(27.59\) and \(52.34\).
04

Making the Decision

Since the test statistic \(40.069\) lies within the critical values \(27.59\) and \(52.34\), we fail to reject the null hypothesis at the \(0.05\) level.

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Key Concepts

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

Standard Deviation
The standard deviation is a measure of how much individual data points differ from the mean of a data set. It gives an idea of the variance or spread of a set of values. When the standard deviation is small, the values are close to the mean. In contrast, a large standard deviation indicates that the values are spread out over a wider range.
Standard deviation is calculated as the square root of the variance. Specifically, for population variance \(\sigma^2\), the standard deviation is \(\sigma = \sqrt{\sigma^2}\). In hypothesis testing, standard deviation plays a crucial role in understanding variability within the data and deciding whether that variability is within the expected range based on the null hypothesis. In our example, we are interested in testing if the standard deviation differs from 12, which corresponds to the variance of 144 because \(12^2 = 144\).
By establishing a null hypothesis about the standard deviation, we can statistically assess if a sample exhibits significantly different variability then what is assumed under the null hypothesis.
Chi-Square Distribution
The chi-square distribution is essential when testing hypotheses about variance. Unlike other distributions like the normal distribution, the chi-square distribution is asymmetrical and only deals with non-negative values since it's based on squared differences. It is primarily used in situations where sample variances are being measured, such as our test here that compares sample variance against a specified population variance.
The shape of the chi-square distribution changes depending on the number of degrees of freedom associated with your sample, \(df = n - 1\), where \(n\) is the sample size. In our problem, the degrees of freedom are 39, because there are 40 samples in the study.
Using the chi-square test statistic \(X^2\), calculated as \((n-1)\times\frac{s^2}{\sigma^2}\), where \(s^2\) is the sample variance, and \(\sigma^2\) is the variance under the null hypothesis, allows us to understand how well our observed sample variance fits the expected variance.
Significance Level
The significance level, often denoted by \(\alpha\), is the probability of rejecting the null hypothesis when it is actually true. It represents the threshold for determining whether an observed effect can be considered statistically significant, in the context of hypothesis testing.
In hypothesis testing, the choice of a significance level, such as 0.05, specifies the risk of falsely claiming a significant difference when none exists, which is known as a Type I error. In our example, we use a 0.05 significance level for a two-tailed test. This implies that there is a 5% risk of concluding that there is a discrepancy in standard deviation when the null hypothesis is correct.
Critical values are derived from the chi-square distribution for the given degrees of freedom and significance level. In this case, the critical values decide the range within which the test statistic must fall to accept the null hypothesis. If our test statistic falls outside this range, we reject the null hypothesis; otherwise, we fail to reject it.
Sample Variance
Sample variance is a measure used to quantify the amount of variation within a sample data set. Calculated as the average of the squared differences from the mean, it provides insights into how much the scores in a sample deviate from the mean. The formula for sample variance \(s^2\) is \(\frac{\sum (x_i - \bar{x})^2}{n-1}\), where \(\bar{x}\) is the sample mean, \(x_i\) represents each value in the sample, and \(n\) is the sample size.
In our problem, the given sample variance is 155. This calculated variance is then used to test against the hypothesized population variance, \(\sigma^2 = 144\).
An interesting aspect of working with sample variance is its use in constructing a test statistic for hypothesis testing. By comparing the sample variance to the given population variance through a predefined statistical model such as the chi-square distribution, we determine the likelihood of observing a sample variance of this extent if the null hypothesis were true.

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Most popular questions from this chapter

One of the objectives of a large medical study was to estimate the mean physician fee for cataract removal. For 25 randomly selected cases, the mean fee was found to be \(3550\)dollar with a standard deviation of \(275\)dollar. Set a \(99 \%\) confidence interval on \(\mu,\) the mean fee for all physicians. Assume fees are normally distributed.

The owner of the Pizza Shack in Exercises 9.177 and 9.178 does not understand the use of the normal distribution and \(z\) in Exercise \(9.178 .\) Help the manager interpret the meaning of the results by redoing both hypothesis tests using \(x=\) number of customers preferring the new crust as the test statistic and its binomial probability distribution. Use a one-tailed test with \(H_{a}: p>0.50\) and \(\alpha=0.02\). The results were as follows: on Tuesday afternoon, they sampled 15 customers and 9 preferred the new pizza crust; on Friday evening, they sampled 200 customers and found 120 preferred the new pizza crust. a. Is there sufficient evidence to conclude a significant preference for the new crust based on Tuesday's customers? b. Is there sufficient evidence to conclude a significant preference for the new crust based on Friday's customers? c. Explain the relationship between the solutions obtained in Exercise 9.178 and here.

Use a computer or calculator to complete the calculations and the hypothesis test for this exercise. Delco Products, a division of General Motors, produces commutators designed to be \(18.810 \mathrm{mm}\) in overall length. (A commutator is a device used in the electrical system of an automobile.) The following data are the lengths of a sample of 35 commutators taken while monitoring the manufacturing process: $$\begin{array}{lllllll}\hline 18.802 & 18.810 & 18.780 & 18.757 & 18.824 & 18.827 & 18.825 \\\18.809 & 18.794 & 18.787 & 18.844 & 18.824 & 18.829 & 18.817 \\\18.785 & 18.747 & 18.802 & 18.826 & 18.810 & 18.802 & 18.780 \\ 18.830 & 18.874 & 18.836 & 18.758 & 18.813 & 18.844 & 18.861 \\\18.824 & 18.835 & 18.794 & 18.853 & 18.823 & 18.863 & 18.808 \\\\\hline\end{array}$$ Is there sufficient evidence to reject the claim that these parts meet the design requirement "mean length is \(18.810 "\) at the \(\alpha=0.01\) level of significance?

Although most people are aware of minor dehydration symptoms such as dry skin and headaches, many are less knowledgeable about the causes of dehydration. According to a poll done for the Nutrition Information Center, the results of a random sample of 3003 American adults showed that \(20 \%\) did not know that caffeine dehydrates. The survey listed a margin of error of plus or minus \(1.8 \%\). a. Describe how this survey of 3003 American adults fits the properties of a binomial experiment. Specifically identify \(n,\) a trial, success, \(p,\) and \(x .\) b. What is the point estimate for the proportion of all Americans who did not know that caffeine dehydrates? Is it a parameter or a statistic? c. Calculate the \(95 \%\) confidence maximum error of estimate for a binomial experiment of 3003 trials that result in an observed proportion of \(0.20 .\) d. How is the maximum error, found in part c, related to the \(1.8 \%\) margin of error quoted in the survey report? e. Find the \(95 \%\) confidence interval for the true proportion \(p\) based on a binomial experiment of 3003 trials that results in an observed proportion of 0.20.

a. What is the relationship between \(p=P(\text { success })\) and \(q=P(\text { failure }) ?\) Explain. b. Explain why the relationship between \(p\) and \(q\) can be expressed by the formula \(q=1-p\) c. If \(p=0.6,\) what is the value of \(q ?\) d. If the value of \(q^{\prime}=0.273,\) what is the value of \(p^{\prime} ?\)

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