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√)

Researchers want to estimate the mean time people spend grocery shopping per week. If they want the margin of error to be within 1.5 hour, how many people should they survey? Assume that they want to calculate a 95% confidence interval and the population standard deviation is about 5.8 hours.

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?

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

A study found that between 2011 and 2015, about 60.1% of workers commuted less than 26 minutes to work. The margin of error for the study was 1.9%.  Find the 95% confidence interval for this situation.(Enter the smallest value in the first blank and the largest value in the last blank. Do not add % sign)( %, % )

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