# The grade point averages for 10 randomly selected junior college students are 2.0, 3.2, 1.8, 2.9, 0.9, 4.0, 3.3, 2.9, 3.6, 0.8. Assume the grade point averages are normally distributed. What is the 98% confidence interval for the true mean?

One way to calculate the confidence interval of a sample mean is to use the student's t-distribution to find that with

One way to calculate the confidence interval of a sample mean is to use the student's t-distribution where the limits of the interval are given by:

where

First we need the sample mean and sample standard deviation:

and the sample standard deviation:

Plugging these values into our first equation we can calculate the limits of the confidence interval for our mean:

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The 98% confidence interval for the true mean of the grade point averages can be calculated using the formula:

[ \text{Confidence Interval} = \bar{x} \pm Z \left( \frac{s}{\sqrt{n}} \right) ]

Where:

- ( \bar{x} ) is the sample mean,
- ( s ) is the sample standard deviation,
- ( n ) is the sample size, and
- ( Z ) is the Z-score corresponding to the desired confidence level.

Given the dataThe 98% confidence interval for the true mean of the grade point averages can be calculated using the formula:

[ \text{Confidence Interval} = \bar{x} \pm Z \left( \frac{s}{\sqrt{n}} \right) ]

Where:

- ( \bar{x} ) is the sample mean,
- ( s ) is the sample standard deviation,
- ( n ) is the sample size, and
- ( Z ) is the Z-score corresponding to the desired confidence level.

Given the

- Sample mean (( \bar{x} )) = 2.74
- Sample standard deviation (( s )) ≈ 1.148
- Sample size (( n )) = 10

From the Z-table or using statistical software, the Z-score for a 98% confidence level is approximately 2.33.

Substituting the values into the formula:

[ \text{Confidence Interval} = 2.74 \pm 2.33 \left( \frac{1.148}{\sqrt{10}} \right) ]

Computing the expression inside the parentheses and then adding and subtracting it from the sample mean will give the confidence interval:

[ \text{Confidence Interval} = 2.74 \pm 2.33 \left( \frac{1.148}{\sqrt{10}} \right) ]

[ \text{Confidence Interval} = 2.74 \pm 0.845 ]

Therefore, the 98% confidence interval for the true mean of the grade point averages is approximately ( (1.895, 3.585) ).

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