Question

Lifetimes of Wristwatches A random sample of the lifetimes of 23 inexpensive wristwatches has a standard deviation of 4.4 months. Assume the variable is normally distributed. Use the chl-square distribution table to find any chi-square values to three decimal places. Round your final answers to one decimal place Part: 0/2 Part 1 of 2 Find the 80% confidence intervals for the variance and standard deviation.

Answer 1

To find the 80% confidence intervals for the variance and standard deviation, we can use the chi-square distribution table. First, we need to find the critical values from the table. Once we have the critical values, we can calculate the confidence intervals for the variance and standard deviation.

Step-by-step explanation:

To find the 80% confidence intervals for the variance and standard deviation, we can use the chi-square distribution table. First, we need to find the critical values from the table. For an 80% confidence level, there will be a 10% chance in each tail. Since the chi-square distribution is symmetric, we can find the critical values at the 5% and 95% percentiles. Using the degrees of freedom, which is the sample size minus 1, we can find the critical values from the table. Once we have the critical values, we can calculate the confidence intervals for the variance and standard deviation.

Let's say the sample size is 23. The degrees of freedom will be 23 - 1 = 22. From the chi-square distribution table, the critical value at the 5% percentile for 22 degrees of freedom is approximately 32.006. The critical value at the 95% percentile is approximately 41.638. To find the confidence interval for the variance, we can use the formula:

(n - 1) * s^2 / x^2 < variance < (n - 1) * s^2 / y^2

where n is the sample size, s is the sample standard deviation, x is the critical value at the 5% percentile, and y is the critical value at the 95% percentile.

Similarly, to find the confidence interval for the standard deviation, we can use the formula:

sqrt((n - 1) * s^2 / y^2) < standard deviation < sqrt((n - 1) * s^2 / x^2)

where n is the sample size, s is the sample standard deviation, x is the critical value at the 5% percentile, and y is the critical value at the 95% percentile.

Plugging in the values, we get:

(22 * 4.4^2) / 41.638^2 < variance < (22 * 4.4^2) / 32.006^2

sqrt((22 * 4.4^2) / 32.006^2) < standard deviation < sqrt((22 * 4.4^2) / 41.638^2)

Calculating these values yields the 80% confidence intervals for the variance and standard deviation.