Question

the proper operation of typical home appliances requires voltage levels that do not vary much. listed below are 5 voltage levels (in volts) at a random house on 5 different days. 119.7 120 120 119.8 120 the mean of the data set is 119.9; the sample standard deviation is 0.14; if the normality plot is not provided you may assume that the voltages are normally distributed. construct a 90% confidence interval for the variance of all voltages in the house. i. procedure select an answer ii. assumptio one variance $x^{2}$ procedure (s) pop one proportion z procedure wn pop one mean t procedure nown the one mean z procedure responses are both greater than 10 sample size is greater than 30 simple random sample normal population iii. unknown parameter: select an answer iv. point estimate: select an answer (round the answer to 3 decimal places) v. confidence level % and $\alpha=$, also $\frac{\alpha}{2}=$, and $1-\frac{\alpha}{2}=$ critical values: (round the answer to 3 decimal places) left= right= vi. lower bound: (round the answer to 3 decimal places) vii. upper bound: (round the answer to 3 decimal places) viii. interpretation: we are % confident that the true population variance is between and

Explanation:

Step1: Determine the procedure

Since we are constructing a confidence interval for the variance and the population is assumed to be normal, we use the One - variance $\chi^{2}$ procedure.

Step2: Identify the unknown parameter

The unknown parameter is the population variance $\sigma^{2}$.

Step3: Find the point estimate

The point estimate for the population variance is the sample variance $s^{2}$. Given $s = 0.14$, then $s^{2}=(0.14)^{2}=0.0196$.

Step4: Calculate $\alpha$ and its related values

The confidence level is $90\%$, so $\alpha=1 - 0.90 = 0.10$. Then $\frac{\alpha}{2}=\frac{0.10}{2}=0.05$ and $1-\frac{\alpha}{2}=1 - 0.05=0.95$.

Step5: Find the critical values

The degrees of freedom $df=n - 1$, where $n = 5$. So $df=5 - 1 = 4$.
Using the $\chi^{2}$ - distribution table or a calculator, the left - critical value $\chi_{1-\frac{\alpha}{2},df}^{2}=\chi_{0.95,4}^{2}=0.711$ and the right - critical value $\chi_{\frac{\alpha}{2},df}^{2}=\chi_{0.05,4}^{2}=9.488$.

Step6: Calculate the lower bound

The formula for the lower bound of the confidence interval for the variance is $\frac{(n - 1)s^{2}}{\chi_{\frac{\alpha}{2},df}^{2}}$.
Substitute $n = 5$, $s^{2}=0.0196$, and $\chi_{\frac{\alpha}{2},df}^{2}=9.488$:
$\frac{(5 - 1)\times0.0196}{9.488}=\frac{4\times0.0196}{9.488}=\frac{0.0784}{9.488}\approx0.008$.

Step7: Calculate the upper bound

The formula for the upper bound of the confidence interval for the variance is $\frac{(n - 1)s^{2}}{\chi_{1-\frac{\alpha}{2},df}^{2}}$.
Substitute $n = 5$, $s^{2}=0.0196$, and $\chi_{1-\frac{\alpha}{2},df}^{2}=0.711$:
$\frac{(5 - 1)\times0.0196}{0.711}=\frac{4\times0.0196}{0.711}=\frac{0.0784}{0.711}\approx0.110$.

Answer:

• Procedure: One variance $\chi^{2}$ procedure
• Unknown parameter: $\sigma^{2}$ (population variance)
• Point estimate: $0.020$
• Confidence level: $90\%$, $\alpha = 0.10$, $\frac{\alpha}{2}=0.05$, $1-\frac{\alpha}{2}=0.95$
• Critical values: left = $0.711$, right = $9.488$
• Lower bound: $0.008$
• Upper bound: $0.110$
• Interpretation: We are $90\%$ confident that the true population variance is between $0.008$ and $0.110$.