QuestionFind a $99\%$ confidence interval for the standard deviation $\sigma$ of washing machine replacement times from a sample of 20, with mean 11.6 and SD 2.2. Calculate $\chi_{L}^{2}$ , $\chi_{R}^{2}$ , and interpret the interval.

Studdy Solution

STEP 1

Assumptions1. The sample size (n) is20. The sample mean (x̄) is11.6 years3. The sample standard deviation (s) is. years4. We are constructing a99% confidence interval for the population standard deviation (σ)
5. We are assuming the replacement times follow a normal distribution

STEP 2

We need to find the chi-square values, $\chi_{}^{2}$ and $\chi_{R}^{2}$ , which are the critical values for the chi-square distribution at the lower and upper ends of the confidence interval. The chi-square distribution is used when estimating the standard deviation of a normally distributed population in cases where the sample size is small and the population standard deviation is unknown.
The chi-square values are found using the formula $\chi^{2} = \frac{(n-1)s^{2}}{\sigma^{2}}$

STEP 3

For a99% confidence interval, the level of significance (α) is1 -0.99 =0.01. Because the confidence interval is two-tailed, we need to divide α by2. So, α/2 =0.005.

STEP 4

The degrees of freedom (df) for a chi-square distribution is n -1. In this case, df =20 -1 =19.

STEP 5

Now, we can find the chi-square values, $\chi_{}^{2}$ and $\chi_{R}^{2}$ , using a chi-square table or a statistical calculator. The lower chi-square value, $\chi_{}^{2}$ , corresponds to the critical value at α/2 =0.005 with19 degrees of freedom. The upper chi-square value, $\chi_{R}^{2}$ , corresponds to the critical value at1 - α/2 =0.995 with19 degrees of freedom.

STEP 6

Next, we will find the confidence interval for the population standard deviation (σ). The confidence interval for σ is given by the formula $\sqrt{\frac{(n-1)s^{2}}{\chi_{R}^{2}}} \leq \sigma \leq \sqrt{\frac{(n-1)s^{2}}{\chi_{}^{2}}}$

STEP 7

Plug in the values for n, s, $\chi_{}^{2}$ , and $\chi_{R}^{2}$ to calculate the confidence interval for σ.

STEP 8

Finally, interpret the confidence interval. The interpretation of a99% confidence interval is that we are99% confident that the true population standard deviation (σ) falls within this range.

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QuestionFind a $99\%$ confidence interval for the standard deviation $\sigma$ of washing machine replacement times from a sample of 20, with mean 11.6 and SD 2.2. Calculate $\chi_{L}^{2}$ , $\chi_{R}^{2}$ , and interpret the interval.

Studdy Solution

STEP 1

Assumptions1. The sample size (n) is20. The sample mean (x̄) is11.6 years3. The sample standard deviation (s) is. years4. We are constructing a99% confidence interval for the population standard deviation (σ)
5. We are assuming the replacement times follow a normal distribution

STEP 2

STEP 3

For a99% confidence interval, the level of significance (α) is1 -0.99 =0.01. Because the confidence interval is two-tailed, we need to divide α by2. So, α/2 =0.005.

STEP 4

The degrees of freedom (df) for a chi-square distribution is n -1. In this case, df =20 -1 =19.

STEP 5

STEP 6

Next, we will find the confidence interval for the population standard deviation (σ). The confidence interval for σ is given by the formula $\sqrt{\frac{(n-1)s^{2}}{\chi_{R}^{2}}} \leq \sigma \leq \sqrt{\frac{(n-1)s^{2}}{\chi_{}^{2}}}$

STEP 7

Plug in the values for n, s, $\chi_{}^{2}$ , and $\chi_{R}^{2}$ to calculate the confidence interval for σ.

STEP 8

Finally, interpret the confidence interval. The interpretation of a99% confidence interval is that we are99% confident that the true population standard deviation (σ) falls within this range.

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QuestionFind a 99%99\%99% confidence interval for the standard deviation σ\sigmaσ of washing machine replacement times from a sample of 20, with mean 11.6 and SD 2.2. Calculate χL2\chi_{L}^{2}χL2​, χR2\chi_{R}^{2}χR2​, and interpret the interval.

Studdy Solution

STEP 1

Assumptions1. The sample size (n) is20. The sample mean (x̄) is11.6 years3. The sample standard deviation (s) is. years4. We are constructing a99% confidence interval for the population standard deviation (σ)5. We are assuming the replacement times follow a normal distribution

STEP 2

STEP 3

For a99% confidence interval, the level of significance (α) is1 -0.99 =0.01. Because the confidence interval is two-tailed, we need to divide α by2. So, α/2 =0.005.

STEP 4

The degrees of freedom (df) for a chi-square distribution is n -1. In this case, df =20 -1 =19.

STEP 5

STEP 6

STEP 7

Plug in the values for n, s, χ2\chi_{}^{2}χ2​, and χR2\chi_{R}^{2}χR2​ to calculate the confidence interval for σ.

STEP 8

Finally, interpret the confidence interval. The interpretation of a99% confidence interval is that we are99% confident that the true population standard deviation (σ) falls within this range.

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Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionFind a $99\%$ confidence interval for the standard deviation $\sigma$ of washing machine replacement times from a sample of 20, with mean 11.6 and SD 2.2. Calculate $\chi_{L}^{2}$ , $\chi_{R}^{2}$ , and interpret the interval.

Assumptions1. The sample size (n) is20. The sample mean (x̄) is11.6 years3. The sample standard deviation (s) is. years4. We are constructing a99% confidence interval for the population standard deviation (σ)
5. We are assuming the replacement times follow a normal distribution

Plug in the values for n, s, $\chi_{}^{2}$ , and $\chi_{R}^{2}$ to calculate the confidence interval for σ.