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In: Statistics and Probability

Assume a normal population with known variance σ2, a random sample (n< 30) is selected. Let...

Assume a normal population with known variance σ2, a random sample (n< 30) is selected. Let x¯,s represent the sample mean and sample deviation. (1)(2pts) write down the formula: 98% one-sided confidence interval with upper bound for the

population mean. (2)(6pts) show how to derive the confidence interval formula in (1).

Expert Solution

Let X has a normal distribution with mean and variance . Then we know that has a Normal distribution with mean and Variance . Therefore by definition we have

will have a N(0,1).

The 98% one-sided confidence interval for is .

2. In general, we know that the confidence interval for a population mean is

This will give a two sided interval. This shows that

Here we are asked to find a one-sided 98% confidence Interval. Since here 98% is greater than 50%, the interval therefore cover the negative side or positive side full, ie one of the limits will either will be (For the left side interval) or for the positive side.

Case i. One sided (left side): Since this is left side, this will start from . We know that

P(. Since 98% is 0.5+0.48. his can be found from EXCEL function NORM.S.INV(0.98) which is 2.0567. Therefore corresponding to 0.48, we have the Z value is 2.0547. ie .

Hence the one sided 98% interval is

Case(ii): If a right sided confidence is considered then it is

orchestra answered 1 year ago

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