Question

The ages of a random sample of five university professors are 39, 54, 61, 72, and 59. Using this information, find a 99% confidence interval for the population standard

Answer 1

Answer: (38.89, 2752.38)

Step-by-step explanation: the confidence interval for population standard deviation is given by the formulae below.

Lower limit

√{(n-1)s²/(X)²α/2}

Upper limit

√{(n-1)s²/(X)²1 -α/2}

Where n = sample size = 5

s² = sample variance.

(X)²α/2 = chi square test statistics value at α/2 level of significance.

(X)²1-α/2 = chi square test statistics value at 1-α/2 level of significance.

The question is telling us to construct a 99% confidence interval, hence our level of significance (α) is 1% = 0.01.

Firstly, we need to get our sample variance (s²).

The formulae for getting sample variance is given below as.

s² = {Σx² - (Σx)²/n}/n - 1

Our table is given below

x : 39, 54, 61, 72, 59......Σx = 285

x² : 1521, 2916, 3721, 5184, 3481......Σx² = 16823

s² = {16823 - (285)²/5}/4

s² = (16823- 16245) / 4

s² = 578/4 = 144.5

From the chi distribution table,

(X)²α/2 = (X)²0.005 = 14.860

(X)²1-α/2 = (X)²0.995 = 0.21

Lower limit

√{(n-1)s²/(X)²α/2 = √{(5-1)×144.5}/14.860

= 4×144.5/14.860 = 578/14.860 = 38.89.

Upper limit

√{(n-1)s²/(X)²1 -α/2} =√(5-1)×144.5/0.21

= 4× 144.5/0.21 = 578/0.21 = 2752.38.

Answer 2

Answer:

$38.898 \leq \sigma^2 \leq 2792.356$

Now we just take square root on both sides of the interval and we got:

$6.237 \leq \sigma \leq 52.843$

Step-by-step explanation:

Data given and notation

34,59,61,71,59

We can calculate the sample standard deviation with the following formula:

$s^2= \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

$s= \sqrt{s^2}$

s=12.021 represent the sample standard deviation

$\bar x$ represent the sample mean

n=5 the sample size

Confidence=99% or 0.99

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi square distribution is the distribution of the sum of squared standard normal deviates .

Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=5-1=4$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.005,4)" "=CHISQ.INV(1-0.005,4)". so for this case the critical values are:

$\chi^2_{\alpha/2}=14.860$

$\chi^2_{1- \alpha/2}=0.207$

And replacing into the formula for the interval we got:

$\frac{(4)(12.021)^2}{14.860} \leq \sigma^2 \leq \frac{(4)(12.021)^2}{0.207}$

$38.898 \leq \sigma^2 \leq 2792.356$

Now we just take square root on both sides of the interval and we got:

$6.237 \leq \sigma \leq 52.843$