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25 votes
25 votes
2) Construct a 99% confidence interval for the population standard deviation σ if a sample of size 23 has standard deviation s=16

User James Khan
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2 Answers

8 votes
8 votes

Final answer:

A 99% confidence interval for the population standard deviation is constructed using the Chi-Square distribution with the sample standard deviation and sample size. The lower and upper limits are computed with the sample size minus one, the sample variance, and the Chi-Square critical values for the desired confidence level.

Step-by-step explanation:

To construct a 99% confidence interval for the population standard deviation σ given a sample standard deviation s of 16 for a sample size of 23, we must use the Chi-Square distribution because the population standard deviation is not known. The degrees of freedom (df) for our sample is df = n - 1 = 23 - 1 = 22. The Chi-Square values that correspond to the 99% confidence level for 22 degrees of freedom can be looked up in a Chi-Square distribution table or calculated using statistical software.

Let's denote the Chi-Square values as χ_12 (lower critical value) and χ_22 (upper critical value). The confidence interval for the population standard deviation is calculated using the formula:

  • L = √((n - 1)s2 / χ_22)
  • U = √((n - 1)s2 / χ_12)

Where L is the lower limit and U is the upper limit of the confidence interval.

We would then plug our values of df, s, and the χ_12 and χ_22 obtained from the Chi-Square distribution into the formula to find the specific interval.

User Mohamed Jakkariya
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2.8k points
4 votes
4 votes

Sample size n = 23

Standard Deviation s = 16

Construct a 99% confidence interval

The expression for the confidence interval is :


\begin{gathered} \text{ Confidence Interval =}\bar{\text{ X}}\pm Z\frac{S}{\sqrt[]{n}}_{} \\ \text{ where X is the }mean,\text{ n is the sameple size, s= standard deviation} \\ Z=Chosen\text{ z value from the table} \end{gathered}

Substitute the value :

Z at the 99% level is Z = 2.576


\begin{gathered} \text{ Confidence Interval =}\bar{\text{ X}}\pm Z\frac{S}{\sqrt[]{n}}_{} \\ \text{ Confidence Interval=}\bar{\text{X}}\text{ }\pm(2.576)\frac{16}{\sqrt[]{23}} \\ \text{Confidence Interval=}\bar{\text{X}}\text{ }\pm8.594 \\ \text{Confidence Interval=(}\bar{\text{X}}\text{ +8.594),(}\bar{\text{X}}-8.594) \end{gathered}

Answer :


\text{Confidence Interval=(}\bar{\text{X}}\text{ +8.594),(}\bar{\text{X}}-8.594)

User Korylprince
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