The sugar content of the syrup in canned peaches is normally distributed. A random sample of n = 10 cans yields a sample standard deviation of s = 4.798 milligrams. Construct a 95% two-sided confidence interval for σ. Round your answers to 2 decimal places.

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Answer:

The 95% confidence interval is given by:

3.30<σ<8.76

Step-by-step explanation:

1) Data given and notation

s=4.798 represent the sample standard deviation

[tex]\bar x[/tex] represent the sample mean

n=10 the sample size

Confidence=95% or 0.95

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]

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

[tex]df=n-1=10-1=9[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a tabel to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.025,9)" "=CHISQ.INV(0.975,9)". so for this case the critical values are:

[tex]\chi^2_{\alpha/2}=19.022[/tex]

[tex]\chi^2_{1- \alpha/2}=2.700[/tex]

And replacing into the formula for the interval we got:

[tex]\frac{(9)(4.798)^2}{19.022} \leq \sigma \frac{(9)(4.798)^2}{2.700}[/tex]

[tex] 10.892 \leq \sigma^2 \leq 76.736[/tex]

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

[tex] 3.30 \leq \sigma \leq 8.76[/tex]

So the best option would be:

3.30<σ<8.76

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