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Bright-Lite claims that its 60 -watt light bulb burns with a length of life that is approximately normally distributed with a standard deviation of 81 hours. A sample of 101 bulbs had a variance of \(8075 .\) Is this sufficient evidence to reject Bright-Lite's claim in favor of the alternative,"the standard deviation is larger than 81 hours," at the 0.05 level of significance?

Short Answer

Expert verified
Based on the steps above, if the calculated test statistic is greater than the critical value, there is sufficient evidence to reject Bright-Lite's claim in favor of the alternative at the 0.05 significance level. If not, we cannot reject Bright-Lite's claim.

Step by step solution

01

State the Null and Alternate Hypotheses

Firstly, define the null and alternative hypotheses. The null hypothesis (\(H_0\)) is Bright-Lite's claim that the standard deviation is 81 hours, or mathematically \(\sigma^2 = 81^2\). The alternative hypothesis (\(H_1\)) is that the standard deviation is larger than 81 hours, or mathematically \(\sigma^2 > 81^2\).
02

Compute Test Statistic

Next, calculate the test statistic using the formula \(\chi^2 = (n-1)s^2 / \sigma_0^2 \) where n is sample size, \(s^2\) is the sample variance and \(\sigma_0^2\) is the hypothesized variance. Here, n=101, \(s^2=8075\) and \(\sigma_0^2 = 81^2\).
03

Determine the Rejection Region

Since this is a right tailed test, we reject the null hypothesis if our test statistic falls in the region \(\chi^2_{0.05,df}\), where df = n-1 = 101 - 1 = 100. Using Chi-Square table, we can find this critical value.
04

Compare Test Statistic with Rejection Region

Next compare the calculated test statistic with the critical value. If the calculated value lies in the rejection region, we reject the null hypothesis.
05

Make Conclusion based on the Test

Finally, if the test statistic falls in the rejection region, then there is sufficient evidence to reject the null hypothesis in favor of the alternative at the 0.05 level of significance. If not, we fail to reject the null hypothesis.

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Key Concepts

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

Understanding Null and Alternative Hypotheses
In hypothesis testing, we start by stating two opposing hypotheses: the null and the alternative. The null hypothesis, denoted as \(H_0\), is a statement of no effect or status quo. It is the assumption that we aim to test. In this case, Bright-Lite's claim sets the null hypothesis as the standard deviation being exactly 81 hours. This translates into a mathematical expression as \(\sigma^2 = 81^2\).

On the other hand, the alternative hypothesis, \(H_1\), challenges the null hypothesis. It suggests that the true situation is different from the claim. Here, it asserts that the standard deviation exceeds 81 hours, captured as \(\sigma^2 > 81^2\).

It's important to notice that the null hypothesis is always tested with the intention to reject it, in favor of the alternative. Essentially, a null hypothesis is only considered true until evidence suggests otherwise. Hypothesis testing allows us to make decisions based on data in the presence of uncertainty, quantifying the evidence through statistical analysis.
The Chi-Square Test for Variance
A Chi-Square test is a statistical method used to test hypotheses about population variances. In the example of the Bright-Lite bulbs, we are particularly interested in whether the variance of bulb lifetimes differs from the claimed variance.

The Chi-Square test works using a test statistic calculated by the formula:
  • \(\chi^2 = \frac{(n-1) \times s^2}{\sigma_0^2} \)
Where \(n\) is the sample size, \(s^2\) is the sample variance, and \(\sigma_0^2\) is the variance according to the null hypothesis.

The value of this test statistic is compared against a critical value from the Chi-Square distribution table to determine whether to reject the null hypothesis. This decision is based on the rejection region, specific to the chosen level of significance (e.g., 0.05).

The use of this test is ideal when examining claims about variances, such as whether the company's claim that bulb lifetimes follow a specific standard deviation holds true or not.
Statistical Significance and Making a Decision
Statistical significance helps us decide whether to reject the null hypothesis in favor of the alternative. It essentially asks: "Is there enough evidence beyond random chance?"

Significance levels are denoted by \(\alpha\), often set at 0.05, and guide our decision boundaries. If the test statistic falls into the predefined rejection region set by \(\alpha\), we say the results are statistically significant. This means it is highly unlikely the observed data would occur if the null hypothesis were true.

In the case of the Bright-Lite bulbs, if the computed Chi-Square statistic exceeds the critical value corresponding to the 0.05 level and 100 degrees of freedom, we consider the results significant enough to reject the null hypothesis, indicating the actual standard deviation might indeed be greater than 81 hours.

This method ensures conclusions are not made prematurely or based purely on subjective interpretation but rather grounded in probabilistic evidence provided by the sample data.

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Most popular questions from this chapter

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