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?

Question

Anthony Arnold · Accepted Answer

One way to calculate the confidence interval of a sample mean is to use the student's t-distribution to find that with #98%# confidence that the mean lies in the range

#1.55, 3.53# 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: #m+- t_(1-alpha//2,N-1)s/sqrt(N)# where #m# is the sample mean #alpha# is the residual probability;in our case #alpha = 1-0.98 = 0.02# #s# is the sample standard deviation #N# is the number of samples (#N-1# is the number of degrees of freedom) First we need the sample mean and sample standard deviation: #m=1/N sum_(i=1)^N x_i=2.54# and the sample standard deviation: #s = sqrt(1/(N-1) sum_(i=1)^N(x_i-m)^2) =1.11# Plugging these values into our first equation we can calculate the limits of the confidence interval for our mean: #2.54+- t_(0.99,9)1.11/sqrt(10)# Where we can look up the value of #t_(0.99,9)# in a table or use a function in a software package, like the T.INV function in Microsoft Excel to find the value #2.82# #2.54+-2.82*1.11/sqrt(10)# therefore we can say with #98%# confidence that the mean lies in the range #1.55, 3.53#

Andrew Adams · Answer

undefined 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) $.