Practice Problem for Module 2
A random sample of 50 students’ GPAs reveals a mean GPA of 2.8 with a standard deviation of 0.45.
- Construct a 95% Confidence Interval for the mean GPA of the population.
We can construct a 95% confidence interval for the population mean GPA using the following information:
Sample mean (x̄) = 2.8
Sample standard deviation (s) = 0.45
Confidence level = 95%
Here’s what we need to do:
Since we don’t know the population standard deviation and the sample size is greater than 30, we can use the z-distribution. For a 95% confidence level, the critical value (z) is 1.96.
Calculate the margin of error:
Margin of error (E) represents the amount we need to add and subtract from the sample mean to create the confidence interval. We can calculate it using the formula:
Where:
z is the critical value (1.96)
s is the sample standard deviation (0.45)
n is the sample size (50)
= 0.13
Construct the confidence interval:
The lower and upper bounds define the confidence interval. We can find them using the following formulas:
Lower bound: x̄ – E = 2.8 – 0.13 = 2.67
Upper bound: x̄ + E = 2.8 + 0.13 = 2.93
Therefore, with a 95% confidence level, we can estimate that the actual population mean GPA falls between 2.67 and 2.93.
Interpretation: We can be 95% confident that the average GPA of the entire population from which this sample was drawn lies somewhere between 2.67 and 2.93.
- If we want to be 90% confident and control the maximum error of estimation to be 0.2, how many more students should be added into the given sample?
To calculate the required sample size for a 90% confidence level with a maximum error of estimation of 0.2, we’ll use the formula for the margin of error (ME):
To find the number of students, we will reverse this formula and will get
Since we can’t have a fraction of a student, we must round up to the nearest whole number. Therefore, we need to add 14 more students to the sample to achieve a 90% confidence level with a maximum estimation error of 0.2.
- Would you conclude that the mean GPA is more than 2.5 at a 5% significance level?
We can conduct a one-sample t-test to test whether the mean GPA is more than 2.5 at a 5% level of significance.
Null Hypothesis (H0): Mean GPA = 2.5
Alternative Hypothesis (H1): Mean GPA > 2.5
We’ll set the significance level to 0.05.
If the calculated p-value is less than 0.05, we reject the null hypothesis and conclude that sufficient evidence suggests that the mean GPA is more than 2.5.
We can use the formula for the t-test:
x̄-µ / (
)
Where:
ˉxˉ is the sample mean GPA (given as 2.8).
μ is the hypothesized population mean GPA (2.5 in this case).
s is the sample standard deviation (given as 0.45).
n is the sample size (given as 50).
Substitute the given values, we get
t =4.716
Now, we need to find the p-value associated with this t-value using a t-distribution table
With level of significance df = n−1
= 50-1
With df=n−1=49, the p-value corresponding to t≈4.716 is significantly less than 0.05 (practically zero).
The p-value corresponding to ≈4.716 is significantly less than 0.05 (practically zero).
Therefore, we reject the null hypothesis.
Conclusion: At the 5% significance level, we have sufficient evidence to conclude that the mean GPA is more than 2.5.
Explore MATH 1300 Practice Problem for Module 2 for more information.