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Use the sample information = 43, 0 = 6, n = 13 to calculate the following confidence intervals for 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.) ​

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To calculate the confidence intervals, one must apply the formula involving the sample mean, standard deviation, sample size, and z-scores for the desired confidence level. The intervals change with each level of confidence, becoming wider as the confidence level increases.

To calculate the confidence intervals for the given sample information where the sample mean (\( \overline{x} \)) is 43, the standard deviation (\( \sigma \)) is 6, and the sample size (n) is 13, we use the formula for a confidence interval which is:

\[ \overline{x} \pm (z * \frac{\sigma}{\sqrt{n}}) \]

Where \( z \) is the z-score that corresponds to the desired confidence level. The z-score for a 90% confidence interval is approximately 1.645, for a 95% confidence interval is approximately 1.960, and for a 99% confidence interval, it is approximately 2.576.

90 Percent Confidence Interval

Multiply the standard error (\( \frac{\sigma}{\sqrt{n}} \)) by the z-score (1.645) and subtract it from and add it to the sample mean to get the interval:

\[ 43 \pm (1.645 * \frac{6}{\sqrt{13}}) \]

\[ 43 \pm 2.7374... \]

\[ 40.2626, 45.7374 \] (rounded to four decimal places)

95 Percent Confidence Interval

Now, using the z-score for 95% confidence which is 1.960:

\[ 43 \pm (1.960 * \frac{6}{\sqrt{13}}) \]

\[ 43 \pm 3.2572... \]

\[ 39.7428, 46.2572 \] (rounded to four decimal places)

99 Percent Confidence Interval

Finally, with the z-score for 99% confidence which is 2.576:

\[ 43 \pm (2.576 * \frac{6}{\sqrt{13}}) \]

\[ 43 \pm 4.2835... \]

\[ 38.7165, 47.2835 \] (rounded to four decimal places)

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