An electric device delivers a current of [tex]$15.0 A$[/tex] for 30 seconds. How many electrons flow through it?
Answer (1)
Calculate the sample standard deviation s = 0.2263 and sample variance s 2 = 0.0512 .
Determine the degrees of freedom: df = n − 1 = 9 .
Find the critical values: χ 0.01 2 = 21.6660 and χ 0.99 2 = 2.0879 .
Calculate the 98% confidence interval for the standard deviation: ( 0.1458 , 0.4698 ) .
Explanation
Understand the problem We are given a sample of 10 voltage levels and asked to construct a 98% confidence interval for the standard deviation of all voltage levels. The data is normally distributed.
Calculate sample statistics First, we need to calculate the sample standard deviation ( s ) and sample variance ( s 2 ) from the given data. Using the provided data [123.7, 123.8, 123.4, 123.5, 123.7, 123.2, 123.6, 123.4, 123.2, 123.2], we find that the sample standard deviation s = 0.2263 and the sample variance s 2 = 0.0512 .
Determine degrees of freedom Next, we determine the degrees of freedom ( df ): df = n − 1 = 10 − 1 = 9 .
Find critical chi-square values We need to find the critical values χ α /2 2 and χ 1 − α /2 2 for a 98% confidence level with df = 9 . Since the confidence level is 98%, α = 1 − 0.98 = 0.02 . Thus, we need to find χ 0.01 2 and χ 0.99 2 with 9 degrees of freedom. From the chi-square distribution table or calculator, we find that χ 0.01 2 = 21.6660 and χ 0.99 2 = 2.0879 .
Calculate lower bound Now, we calculate the lower bound of the confidence interval for the standard deviation using the formula: χ α /2 2 ( n − 1 ) s 2 = 21.6660 ( 10 − 1 ) ( 0.0512 ) = 21.6660 9 ( 0.0512 ) = 21.6660 0.4608 = 0.02126 = 0.1458
Calculate upper bound Next, we calculate the upper bound of the confidence interval for the standard deviation using the formula: χ 1 − α /2 2 ( n − 1 ) s 2 = 2.0879 ( 10 − 1 ) ( 0.0512 ) = 2.0879 9 ( 0.0512 ) = 2.0879 0.4608 = 0.2207 = 0.4698
State the confidence interval Therefore, the 98% confidence interval for the standard deviation is (0.1458, 0.4698).
Examples
Understanding the variability in voltage levels is crucial for maintaining the proper functioning of home appliances. For instance, if a 98% confidence interval for the standard deviation of voltage levels is found to be (0.1458, 0.4698) volts, it suggests that the true standard deviation of voltage levels in the home lies within this range with 98% confidence. This information can help homeowners and electricians assess the stability of the electrical supply and take necessary measures to prevent damage to sensitive electronic equipment. For example, if the upper bound of the interval is too high, it may indicate a need for voltage regulation to protect appliances from power surges.
