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Find the 90% confidence interval for the variance and standard deviation for the price in dollars of an adult single-day ski lift ticket. The data represent a selected sample of nationwide ski resorts. Assume the variable is normally distributed. 59 54 53 52 51 39 49 46 49 48

User Wnnmaw
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Final answer:

For the given sample data on adult single-day ski lift ticket prices, the 90% confidence interval for the variance is approximately (12.89, 56.19), and the confidence interval for the standard deviation is approximately (3.59, 7.49).

Step-by-step explanation:

To find the confidence interval for the variance and standard deviation, you can use the chi-squared distribution. For a 90% confidence interval, the degrees of freedom would be n-1, where n is the sample size. In this case, n = 10, so the degrees of freedom would be 9.

  • Compute the sample variance: s² = 24.67
  • Determine the critical values from the chi-squared distribution for a 90% confidence interval with 9 degrees of freedom: Chi-squared values are approximately 14.684 and 21.666.
  • Calculate the confidence interval for the variance: (n-1)s² / Chi-squared upper, (n-1)s² / Chi-squared lower = (9 * 24.67 / 21.666, 9 * 24.67 / 14.684) ≈ (12.89, 56.19).
  • Calculate the confidence interval for the standard deviation by taking the square root of the corresponding variance values: (√12.89, √56.19) ≈ (3.59, 7.49).
User Jared Nedzel
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