The highest temperatures (in degrees Fahrenheit) of a random sample of 7 towns are:

97.9, 98.6, 96.4, 99.4, 99.3, 98.5, 98.9

Assume high temperatures are normally distributed. Based on this data, find the 95% confidence interval of the mean high temperature of the towns. Enter your answer as an open interval (i.e., parentheses) accurate to two decimal places. (The sample data are reported accurate to one decimal place.)

95% C.I. = ?

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

You measure the weights of 29 watermelons and find they have a mean weight of 80 ounces. Assume the population standard deviation is 2.8 ounces. Based on this, what is the maximal margin of error associated with a 99% confidence interval for the true population mean watermelon weight?

Give your answer as a decimal, to two decimal places: ± ______ ounces.

The 95% confidence interval of the mean high temperature of towns is (97.43, 99.71) degrees Fahrenheit. The maximal margin of error associated with a 99% confidence interval for the true population mean watermelon weight is ±1.04 ounces.

1. To find the 95% confidence interval of the mean high temperature of towns, we can use the formula:
Confidence Interval = Mean ± (Critical Value) * (Standard Deviation / √(Sample Size))
The sample mean of the high temperatures is found by adding up all the temperatures and dividing by the sample size (which is 7):
Sample Mean = (97.9 + 98.6 + 96.4 + 99.4 + 99.3 + 98.5 + 98.9) / 7 = 98.57 (rounded to two decimal places)

The standard deviation measures the spread of the data. To calculate it, we need to find the differences between each temperature and the sample mean, square those differences, sum them up, divide by the sample size minus 1, and then take the square root:
Standard Deviation = √[((97.9 - 98.57)² + (98.6 - 98.57)² + (96.4 - 98.57)² + (99.4 - 98.57)² + (99.3 - 98.57)² + (98.5 - 98.57)² + (98.9 - 98.57)²) / (7 - 1)]
Standard Deviation ≈ 1.06 (rounded to two decimal places)

The critical value is based on the desired confidence level. For a 95% confidence level, the critical value is 1.96 (rounded to three decimal places).

The margin of error represents the maximum amount by which the sample mean can differ from the true population mean. It is found by multiplying the critical value by the standard deviation divided by the square root of the sample size:
Margin of Error = (Critical Value) * (Standard Deviation / √(Sample Size))
Margin of Error ≈ (1.96) * (1.06 / √(7))
Margin of Error ≈ 1.14 (rounded to two decimal places)

The confidence interval is found by subtracting the margin of error from the sample mean and adding it to the sample mean:
Confidence Interval = (Sample Mean - Margin of Error, Sample Mean + Margin of Error)
Confidence Interval ≈ (98.57 - 1.14, 98.57 + 1.14)
Confidence Interval ≈ (97.43, 99.71) (rounded to two decimal places)

2. To find the maximal margin of error associated with a 99% confidence interval for the true population mean watermelon weight, we can use the formula:
Margin of Error = (Critical Value) * (Standard Deviation / √(Sample Size))
Given that the sample mean weight is 80 ounces, the population standard deviation is 2.8 ounces, and the sample size is 29, we need to find the critical value for a 99% confidence level.

For a 99% confidence level, the critical value is 2.576 (rounded to three decimal places).

The margin of error is found by multiplying the critical value by the standard deviation divided by the square root of the sample size:
Margin of Error = (Critical Value) * (Standard Deviation / √(Sample Size))
Margin of Error = (2.576) * (2.8 / √(29))
Margin of Error ≈ 1.036 (rounded to three decimal places)

Therefore, the maximal margin of error associated with a 99% confidence interval for the true population mean watermelon weight is ±1.04 ounces.

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