Students enrolled in an introductory statistics course at a university were asked to take a survey that indicated whether the student has a visual or verbal learning style. Of the 39 students who took the survey, 25 were judged to have a visual learning style and 14 were considered verbal learners.
1. Determine a 90% confidence interval for the population proportion who are visual learners at this university.
2. Write a sentence interpreting what this interval says.
3. How would a 99% confidence interval compare to this one in terms of its midpoint and half-width? (Do not bother to determine this interval.)
4. Check whether the technical condition concerning sample size is satisfied here.
5. Explain why you might feel wary about applying this confidence interval to the population of all students at this university.
1. To determine a 90% confidence interval for the population proportion who are visual learners, you can use the formula:
Confidence Interval = Sample Proportion ± Margin of Error
First, calculate the sample proportion (p-hat) by dividing the number of students with a visual learning style by the total number of students who took the survey:
p-hat = 25 / 39 ≈ 0.641
Next, calculate the margin of error using the formula:
Margin of Error = Critical Value * Standard Error
The critical value can be found using a standard normal distribution table or a calculator. For a 90% confidence level, the critical value is approximately 1.645.
The standard error is calculated using the formula:
Standard Error = √(p-hat * (1 - p-hat) / n)
where n is the sample size.
Substituting the values into the formula, you get:
Standard Error = √(0.641 * (1 - 0.641) / 39) ≈ 0.081
Finally, calculate the confidence interval:
Confidence Interval = 0.641 ± 1.645 * 0.081
2. The 90% confidence interval for the population proportion who are visual learners is approximately 0.641 ± 0.133.
This means that we are 90% confident that the true proportion of visual learners at this university falls between 0.508 and 0.774.
3. A 99% confidence interval would have a larger margin of error and therefore a greater half-width. The midpoint of the interval would likely be similar to the 90% confidence interval, but the range of values would be wider to provide a higher level of confidence.
4. The technical condition concerning sample size is satisfied if the number of successes (students with a visual learning style) and failures (students with a verbal learning style) are both greater than 5. In this case, we have 25 successes and 14 failures, which satisfies the condition.
5. There are a few reasons why you might feel wary about applying this confidence interval to the population of all students at this university.
First, the survey was only given to students enrolled in an introductory statistics course, which may not be representative of the entire student population. There could be specific characteristics or differences among these students that may not be present in the broader population.
Second, the sample size of 39 may be relatively small for making generalizations about the entire university population. A larger sample size would provide greater precision and accuracy in estimating the true proportion.
Lastly, the survey relies on self-reported learning styles, which may introduce bias or inaccuracies. Some students may have misjudged their learning style or provided incomplete or incorrect information.
Taking these concerns into account, it is important to interpret the confidence interval cautiously and consider the limitations of the survey and sample when applying it to the larger population.
1. To determine a 90% confidence interval for the population proportion who are visual learners at this university, we can use the formula for estimating the proportion.
The point estimate for the population proportion of visual learners is: p̂ = x/n, where x is the number of visual learners and n is the total number of students surveyed.
In this case, x = 25 (number of visual learners) and n = 39 (total number of students surveyed).
The formula for the confidence interval is: p̂ ± z * √[(p̂(1-p̂))/n], where z is the z-score corresponding to the desired level of confidence. For a 90% confidence level, the z-score is approximately 1.645.
Plugging in the values, we get: p̂ ± 1.645 * √[(p̂(1-p̂))/n]
Calculating the interval, we have: 25/39 ± 1.645 * √[(25/39)(14/39)/39]
2. The 90% confidence interval for the population proportion of visual learners at this university is (0.535, 0.795). This means that we are 90% confident that the true proportion of visual learners at the university falls between 53.5% and 79.5%.
3. A 99% confidence interval would have a larger z-score than the 90% confidence interval, which means it would have a wider interval. The midpoint of the 99% confidence interval would remain the same as the 90% confidence interval, but the half-width (margin of error) would be larger.
4. The technical condition concerning sample size is satisfied if both np̂ ≥ 10 and n(1-p̂) ≥ 10, where np̂ is the number of successes and n(1-p̂) is the number of failures. In this case, np̂ = 25 and n(1-p̂) = 14. Since both are greater than 10, the technical condition is satisfied.
5. We might feel wary about applying this confidence interval to the population of all students at this university because the sample size is relatively small (39 students). A larger sample size would provide more reliable estimates and reduce the margin of error. Additionally, these results are specific to the students who took the survey and might not be representative of the entire population of students at the university.