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------------------------------------------------ 1. Assume that a sample is used to estimate a population mean \(\mu\). Find the margin of error (M.E.) that corresponds to a sample of size 18 with a mean of 30.8 and a standard deviation of 9.7 at a confidence level of 90%.

- Report M.E. accurate to one decimal place because the sample statistics are presented with this accuracy.
- Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

2. You must estimate the mean temperature (in degrees Fahrenheit) with the following sample temperatures:

- 81.4
- 64.2
- 73.8
- 78.6
- 66.5
- 97.7
- 84.5

Find the 80% confidence interval. Enter your answer as an open interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place).

- 80% C.I. = ?

- Answer should be obtained without any preliminary rounding.

The 80% confidence interval for estimating the mean temperature is approximately (66.43, 89.77), which is an open-interval represented by parentheses.

To find the margin of error (M.E.) for estimating a population mean at a given confidence level, we can use the formula:

M.E. = Critical value * Standard deviation / √(sample size)

Sample size (n) = 18

Sample mean (x(bar)) = 30.8

Standard deviation (σ) = 9.7

Confidence level = 90%

First, we need to find the critical value corresponding to a 90% confidence level. This critical value is associated with the z-score for the desired confidence level.

Using a standard normal distribution table or a calculator, we find that the critical value for a 90% confidence level is approximately 1.645 (rounded to 3 decimal places).

Now, we can calculate the margin of error:

M.E. = 1.645 * 9.7 / √(18)

M.E. ≈ 3.058 (rounded to 3 decimal places)

Therefore, the margin of error (M.E.) for estimating the population mean at a 90% confidence level is approximately 3.058 (rounded to 3 decimal places).

To find the 80% confidence interval for estimating the mean temperature, we can use the formula:

Confidence Interval = Sample mean ± Margin of Error

Given the sample temperatures: 81.4, 64.2, 73.8, 78.6, 66.5, 97.7, 84.5.

First, we need to calculate the sample mean and standard deviation.

Sample mean (x(bar)) = (81.4 + 64.2 + 73.8 + 78.6 + 66.5 + 97.7 + 84.5) / 7 ≈ 78.1

Next, we need to calculate the margin of error (M.E.) using the formula:

M.E. = Critical value * Standard deviation / √(sample size)

The critical value for an 80% confidence level can be found using a t-distribution table or calculator. For a sample size of 7, the critical value is approximately 1.894 (rounded to 3 decimal places).

Standard deviation (s) can be calculated as the square root of the sample variance.

Sample variance = [(81.4 - 78.1)² + (64.2 - 78.1)² + (73.8 - 78.1)² + (78.6 - 78.1)² + (66.5 - 78.1)² + (97.7 - 78.1)² + (84.5 - 78.1)²] / (7-1) ≈ 152.757

Standard deviation (s) = √(152.757) ≈ 12.358

Now, we can calculate the margin of error:

M.E. = 1.894 * 12.358 / √(7)

M.E. ≈ 11.672 (rounded to 3 decimal places)

Finally, we can construct the confidence interval:

Confidence Interval = Sample mean ± Margin of Error

Confidence Interval = 78.1 ± 11.672

Confidence Interval ≈ (66.428, 89.772) (rounded to two decimal places)

Therefore, the 80% confidence interval for estimating the mean temperature is approximately (66.43, 89.77) (rounded to two decimal places), which is an open-interval represented by parentheses.

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