To estimate the average study hours of all students taking QM at Dawson College using the given data set, we need to calculate the margin of error (ME) and the confidence interval bounds at the 90% confidence level. Given: - Sample size (\( n \)) = 53- Mean (\( \bar{x} \)) = 5.434- Standard deviation (\( s \)) = 2.422### Step-by-Step Calculation: #### e) Margin of Error (ME) at the 90% confidence level: 1. **Find the critical value (z\(_{\alpha/2}\)) for a 90% confidence level:** - For a 90% confidence level, \(\alpha = 0.10\), so \(\alpha/2 = 0.05\). - The critical value (z\(_{\alpha/2}\)) for 90% confidence is approximately 1.645. 2. **Calculate the standard error (SE):** \[ SE = \frac{s}{\sqrt{n}} = \frac{2.422}{\sqrt{53}} \approx 0.332 \] 3. **Calculate the margin of error (ME):** \[ ME = z_{\alpha/2} \times SE = 1.645 \times 0.332 \approx 0.546 \] #### f) Lower Bound (LB) at the 90% confidence level: \[ LB = \bar{x} - ME = 5.434 - 0.546 \approx 4.888\] #### g) Upper Bound (UB) at the 90% confidence level: \[ UB = \bar{x} + ME = 5.434 + 0.546 \approx 5.980\] ### Summary of Results: - **Margin of Error (ME):** 0.546- **Lower Bound (LB):** 4.888- **Upper Bound (UB):** 5.980These calculations provide the 90% confidence interval for the average study hours of all students ta

Calculate the margin of error and construct the confidence interval for the population mean using the Student's t-distribution (you may assume the population data is normally distributed).10:41 p.m.x̄ =36.4, n=43, s=17, 98% confidence

After surveying 500 college students, a marketing company found that the average amount of time college students spend doing laundry is 2 hours per week.  The standard deviation is 0.25 hour, and confidence interval is 95%.  (Note: The critical value for a 95% confidence interval is 1.96.)  Use the formula below to find the margin of error, where s is the standard deviation and n is the sample size.  Margin of Error = 1.96(sn√)

A group of statistics students decided to conduct a survey at their university to find the average (mean) amount of time students spent studying per week. Assuming a population standard deviation of twenty three hours, what is the required sample size if the error should be less than three hours with a 95% level of confidence?

A study of a local high school tried to determine the mean number of text messages that each student sent per day.  The study surveyed a random sample of 30 students in the high school and found a mean of 181 messages sent per day with a standard deviation of 70 messages.  At the 95% confidence level, find the margin of error for the mean, rounding to the nearest whole number.  Margin of Error = 2(sn√)2(𝑠𝑛)where s is the standard deviation and n is the sample size.

A random sample of 30 students taking statistics at a community college found the student’s mean GPA to be 3.25. A 95% confidence interval for the mean GPA of all students taking statistics at this college was determined to be (3.07, 3.43). What is the margin of error for this confidence interval? 0.18 0.95 3.25 0.36