Respuesta :
Answer:
[tex] 38.898 \leq \sigma^2 \leq 2792.356[/tex]
Now we just take square root on both sides of the interval and we got:
[tex] 6.237 \leq \sigma \leq 52.843[/tex]
Step-by-step explanation:
Data given and notation
34,59,61,71,59
We can calculate the sample standard deviation with the following formula:
[tex] s^2= \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]
[tex] s= \sqrt{s^2}[/tex]
s=12.021 represent the sample standard deviation
[tex]\bar x[/tex] 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:
[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]
The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:
[tex]df=n-1=5-1=4[/tex]
Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], 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:
[tex]\chi^2_{\alpha/2}=14.860[/tex]
[tex]\chi^2_{1- \alpha/2}=0.207[/tex]
And replacing into the formula for the interval we got:
[tex]\frac{(4)(12.021)^2}{14.860} \leq \sigma^2 \leq \frac{(4)(12.021)^2}{0.207}[/tex]
[tex] 38.898 \leq \sigma^2 \leq 2792.356[/tex]
Now we just take square root on both sides of the interval and we got:
[tex] 6.237 \leq \sigma \leq 52.843[/tex]
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.
