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Chapter 11: Problem 49

The manufacturer of a certain brand of lightbulbs claims that the variance of the lives of these bulbs is 4200 square hours. A consumer agency took a random sample of 25 such bulbs and tested them. The variance of the lives of these bulbs was found to be 5200 square hours. Assume that the lives of all such bulbs are (approximately) normally distributed. a. Make the \(99 \%\) confidence intervals for the variance and standard deviation of the lives of all such bulbs. b. Test at a \(5 \%\) significance level whether the variance of such bulbs is different from 4200 square hours.

Short Answer

Expert verified
The confidence interval for variance and the standard deviation can be calculated using the provided sample data. The hypothesis test lets us test whether the variance of life expectancy is significantly different from 4200 based on our 5% significance level. The exact values will require plugging numbers into the described formulas and calculations.

Step by step solution

01

Calculate Confidence Interval for Variance

Given that \( n = 25 \) and \( s^{2} = 5200 \), we can apply the Chi-Square distribution to calculate the confidence interval of the variance. Using the versions \( \chi^{2}_{\alpha/2, n-1} \) and \( \chi^{2}_{1-\alpha/2, n-1} \), we have the confidence interval as \( \left[ (n-1)s^{2}/\chi^{2}_{0.005,24}, (n-1)s^{2}/\chi^{2}_{0.995,24} \right] \).
02

Translate Variance to Standard Deviation

To find the standard deviation interval, take the square root of the variance interval: \( \left[ \sqrt{(n-1)s^{2}/\chi^{2}_{0.005, 24}}, \sqrt{(n-1)s^{2}/\chi^{2}_{0.995, 24}} \right] \).
03

Hypothesis Testing

Null hypothesis is \( H_{0}: \sigma^{2} = 4200 \), while the alternative hypothesis is \( H_{1}: \sigma^{2} \neq 4200 \). The test statistic is \( \chi^{2} = (n-1)s^{2}/\sigma_{0}^{2} \). Compare this to the critical value from the Chi-Square distribution, \( \chi^{2}_{\alpha/2, n-1} \) and \( \chi^{2}_{1-\alpha/2, n-1} \). If the test statistic falls within this area, we don't reject the null hypothesis. If it lies outside, we reject the null hypothesis.

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Most popular questions from this chapter

A 2010 poll by Marist University asked people to choose their favorite classic Christmas movie from a list of five choices. The following table shows the frequencies for the various movies: $$ \begin{array}{l|ccccc} \hline & \text { It's A } & \text { A Christmas } & \text { Miracle on } & \text { White } & \text { A Christmas } \\ \text { Movie } & \text { Wonderful Life } & \text { Story } & \text { 34th Street } & \text { Christmas } & \text { Carol } \\ \hline \text { Frequency } & 247 & 237 & 226 & 124 & 134 \\ \hline \end{array} $$

A drug company is interested in investigating whether the color of their packaging has any impact on sales. To test this, they used five different colors (blue, green, orange, red, and yellow) for the boxes of an over-the- counter pain reliever, instead of their traditional white box. The following table shows the number of boxes of each color sold during the first month. $$ \begin{array}{l|ccccc} \hline \text { Box color } & \text { Blue } & \text { Green } & \text { Orange } & \text { Red } & \text { Yellow } \\ \hline \text { Number of boxes sold } & 310 & 292 & 280 & 216 & 296 \\ \hline \end{array} $$ Using a \(1 \%\) significance level, test the null hypothesis that the number of boxes sold of each of these five colors. is the same.

The percentage distribution of birth weights for all children in cases of multiple births (twins, triplets, etc.) in North Carolina during 2009 was as given in the following table: $$ \begin{array}{l|cccc} \hline \text { Weight (grams) } & 0-500 & 501-1500 & 1501-2500 & 2501-8165 \\ \hline \text { Percentage } & 1.45 & 11.02 & 49.23 & 38.30 \\ \hline \end{array} $$ The frequency distribution of birth weights of a sample of 587 children who shared multiple births and were born in North Carolina in 2012 is as shown in the following table? $$ \begin{array}{l|cccc} \hline \text { Weight (grams) } & 0-500 & 501-1500 & 1501-2500 & 2501-8165 \\ \hline \text { Frequency } & 2 & 60 & 305 & 220 \\ \hline \end{array} $$ Test at a \(2.5 \%\) significance level whether the 2012 distribution of birth weights for all children born in North Carolina who shared multiple births is significantly different from the one for \(2009 .\)

To make a goodness-of-fit test, what should be the minimum expected frequency for each category? What are the alternatives if this condition is not satisfied?

Explain how the expected frequencies for cells of a contingency table are calculated in a test of independence or homogeneity. How do you find the degrees of freedom for such tests?
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