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Answer:
The 98% confidence interval for the variance in the pounds of impurities would be [tex]8.400 \leq \sigma^2 \leq 39.827[/tex].
Step-by-step explanation:
1) Data given and notation
[tex]s^2 =16[/tex] represent the sample variance
s=4 represent the sample standard deviation
[tex]\bar x[/tex] represent the sample mean
n=20 the sample size
Confidence=98% or 0.98
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 .
2) 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]
On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.
The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:
[tex]df=n-1=20-1=19[/tex]
Since the Confidence is 0.98 or 98%, the value of [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex], and we can use excel, a calculator or a table to find the critical values.
The excel commands would be: "=CHISQ.INV(0.01,19)" "=CHISQ.INV(0.99,19)". so for this case the critical values are:
[tex]\chi^2_{\alpha/2}=36.191[/tex]
[tex]\chi^2_{1- \alpha/2}=7.633[/tex]
And replacing into the formula for the interval we got:
[tex]\frac{(19)(16)}{36.191} \leq \sigma^2 \leq \frac{(19)(16)}{7.633}[/tex]
[tex] 8.400 \leq \sigma^2 \leq 39.827[/tex]
So the 98% confidence interval for the variance in the pounds of impurities would be [tex]8.400 \leq \sigma^2 \leq 39.827[/tex].
The 98% confidence interval for the mean change in score is between 69.92 points to 74.08 points
How to calculate confidence interval
Standard deviation = √variance = √16 = 4
The z score of 98% confidence interval is 2.326
The margin of error (E) is:
[tex]E = Z_\frac{\alpha }{2} *\frac{standard\ deviation}{\sqrt{sample\ size} } =2.326*\frac{4}{\sqrt{20} } =2.08[/tex]
The confidence interval = mean ± E = 72 ± 2.08 = (69.92, 74.08)
The 98% confidence interval for the mean change in score is between 69.92 points to 74.08 points
Find out more on confidence interval at: https://brainly.com/question/15712887
