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Im not sure about the second one but the first one is 1099-DIV and the last one is taxable income. I can tell you the second one is NOT deductions haha

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The ages of a random sample of five university professors are 39, 54, 61, 72, and 59. Using this information, find a 99% confide

blsea [12.9K]

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.