Question

Use the sample information x¯ = 43, σ = 3, n = 13 to calculate the following confidence intervals for μ assuming the sample is from a normal population. (a) 90 percent confidence. (Round your answers to 4 decimal places.) The 90% confidence interval is from to (b) 95 percent confidence. (Round your answers to 4 decimal places.) The 95% confidence interval is from to (c) 99 percent confidence. (Round your answers to 4 decimal places.) The 99% confidence interval is from to (d) Describe how the intervals change as you increase the confidence level. The interval gets narrower as the confidence level increases. The interval gets wider as the confidence level decreases. The interval gets wider as the confidence level increases. The interval stays the same as the confidence level increases.

Answer 1

From the data: , $\bar{x}$ = 43, $\sigma$ = 3, n = 13

The Confidence Interval is given by $\bar{x}$ $\pm$ ME

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(a) 90% Confidence interval

The Zcritical (2 tail) for $\alpha$ = 0.10 is 1.645

$ME = tcritical * \frac{\sigma}{\sqrt{n}} = 1.645 * \frac{3}{\sqrt{13}} = 1.3687$

The Lower Limit = 43 - 1.3687 = 41.6313

The Upper Limit = 43 + 1.3687 = 44.3687

The 90% Confidence Interval is from 41.6313 to 44.3687)

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(b) 95% Confidence interval

The Zcritical (2 tail) for $\alpha$ = 0.05 is 1.96

$ME = tcritical * \frac{\sigma}{\sqrt{n}} = 1.96 * \frac{3}{\sqrt{13}} = 1.6308$

The Lower Limit = 43 - 1.6308 = 41.3692

The Upper Limit = 43 + 1.6308 = 44.6308

The 95% Confidence Interval is from 41.3692 to 44.6308)

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(c) 99% Confidence interval

The Zcritical (2 tail) for $\alpha$ = 0.01 is 2.576

$ME = tcritical * \frac{\sigma}{\sqrt{n}} = 2.576 * \frac{3}{\sqrt{13}} = 2.1434$

The Lower Limit = 43 - 2.1434 = 40.8566

The Upper Limit = 43 + 2.1434 = 45.1434

The 99% Confidence Interval is from 40.8566 to 45.1434)

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(d) In general the interval gets wider as the confidence level increases.

This is because, the critical value is in the numerator, and as the confidence level increases, the critical value increases, thus incfreasing the value of ME. This increase in ME makes the interval wider.

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