Lifetimes of Wristwatches A random sample of the lifetimes of 29 inexpensive wristwatches has a standard deviation of 4.8 months. Assume the variable is normally distributed. Use the chi-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 98% confidence intervals for the variance and standard deviation.

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Section13.3: Special Probability Density Functions
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Lifetimes of Wristwatches Find the 99% confidence
interval for the variance and standard deviation for the
lifetimes of inexpensive wristwatches if a sample of 23
watches has a standard deviation of 4.7 months. Assume the
variable is normally distributed. Use the chi-square
distribution table to find any chi-square values to three
decimal places. Round your final answers to one decimal
place. Do you feel that the lifetimes are relatively
consistent?
Explanation
Formula for the Confidence Interval for a
Variance
(n – 1)s²
(n – 1)s²
Xieft
d.f. =n-1
Since a= 0.01, the two critical values, respectively, for the 0.005
and 0.995 levels for 22 d.f. are
Xnht = 42.796, n = 8.643
Xright
The 99% confidence interval for the variance is found by
substituting in the formula.
(n – 1)s²
(n- 1)s²
<o? <
Xieft
(23 – 1)(4.7)²
< g? <
(23 – 1)(4.7)²
42.796
8.643
11.4 < o? < 56.2
Hence, you can be 99% confident that the true variance for
the lifetimes of inexpensive wristwatches is between 11.4 and
56.2.
For the standard deviation, the confidence interval is
V11.4 < o < 56.2
3.4 < o< 7.5
Hence, you can be 99% confident that the true standard
deviation for the lifetimes of inexpensive wristwatches is
between 3.4 and 7.5 months, based on a sample of 23
wristwatches.
Transcribed Image Text:Lifetimes of Wristwatches Find the 99% confidence interval for the variance and standard deviation for the lifetimes of inexpensive wristwatches if a sample of 23 watches has a standard deviation of 4.7 months. Assume the variable is normally distributed. Use the chi-square distribution table to find any chi-square values to three decimal places. Round your final answers to one decimal place. Do you feel that the lifetimes are relatively consistent? Explanation Formula for the Confidence Interval for a Variance (n – 1)s² (n – 1)s² Xieft d.f. =n-1 Since a= 0.01, the two critical values, respectively, for the 0.005 and 0.995 levels for 22 d.f. are Xnht = 42.796, n = 8.643 Xright The 99% confidence interval for the variance is found by substituting in the formula. (n – 1)s² (n- 1)s² <o? < Xieft (23 – 1)(4.7)² < g? < (23 – 1)(4.7)² 42.796 8.643 11.4 < o? < 56.2 Hence, you can be 99% confident that the true variance for the lifetimes of inexpensive wristwatches is between 11.4 and 56.2. For the standard deviation, the confidence interval is V11.4 < o < 56.2 3.4 < o< 7.5 Hence, you can be 99% confident that the true standard deviation for the lifetimes of inexpensive wristwatches is between 3.4 and 7.5 months, based on a sample of 23 wristwatches.
Lifetimes of Wristwatches A random sample of the lifetimes of 29
inexpensive wristwatches has a standard deviation of 4.8 months. Assume the
variable is normally distributed. Use the chi-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 98% confidence intervals for the variance and standard deviation.
<o<
Transcribed Image Text:Lifetimes of Wristwatches A random sample of the lifetimes of 29 inexpensive wristwatches has a standard deviation of 4.8 months. Assume the variable is normally distributed. Use the chi-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 98% confidence intervals for the variance and standard deviation. <o<
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