5. Jennifer wanted to know the average price of shoes that her customer purchased. She sampled 160 pairs of shoes that were sold and found out that the mean average price is ₱800 with a standard deviation of ₱75. Construct a 95% confidence interval for the mean price of all shoes that were sold. Which is the correct sequence in finding the length of the confidence interval?   I. Compute for the margin of error.   II. Compute for the upper and lower limit of the confidence interval.   III. Write the given data.   IV. Write the confidence interval.V. Compute for the length of the confidence interval.*A. I,II,III,IV,VB. II,I,III,IV,VC. III,I,II,IV,VD. V,IV,III,II,I

2.  The average weight of 100 sold chicken is 3 kilos. The standard deviation of the weights of all the chicken in the store is 1.1 kilos and an 85% level of confidence is to be used. What is the length of the confidence interval?*A. 0.3168B. 0.4268C. 0.51268D. 0.65168

Compute the margin of error at a 95% confidence level. (Round final answer to 2 decimal places.)

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

(c)For the mileage values in this sample, 36.4 thousand miles is more extreme than 25.3 thousand miles is, that is, 36.4 is farther from the sample mean mileage than 25.3 is. How would the 90% confidence interval for the mean used selling price when the mileage is 36.4 thousand miles compare to the 90% confidence interval for the mean used selling price when the mileage is 25.3 thousand miles? The intervals would be identical. The interval computed from a mileage of 36.4 thousand miles would be narrower and have a different center. The interval computed from a mileage of 36.4 thousand miles would be wider but have the same center. The interval computed from a mileage of 36.4 thousand miles would be narrower but have the same center. The interval computed from a mileage of 36.4 thousand miles would be wider and have a different center.

a.Construct a 95%confidence interval estimate of the mean cost of a meal for restaurants that have a summated rating of 51. 19.857≤Y1X=×1≤26.923 (Type integers or decimals.Round to three decimal places as needed.Use ascending order.) b.Construct a 95%prediction interval of the cost of a meal for an individual restaurant that has a summated rating of 51. 4.498≤Yx=51≤42.283 (Type integers or decimals.Round to three decimal places as needed.Use ascending order.) c.Explain the differences in the results from (a)and(b).Choose the correct answer below. OA.Since there is more variation in estimating a mean value than in predicting an individual value,the confidence interval is narrower than the prediction interval,holding everything else fixed. B.Since there is more variation in estimating a mean value than in predicting an individual value,the confidence interval is wider than the prediction interval,holding everything else fixed. O C.Since there is more variation in predicting an individual value than in estimating a mean value,the prediction interval is wider than the confidence interval,holding everything else fixed. OD.Since there is more variation in predicting an individual value than in estimating a mean value,the prediction interval is narrower than the confidence interval,holding everything else fixed.