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.

Use the following information for questions 17 and 18.You randomly select 16 restaurants and measure the temperature of the coffee sold at each. The sample mean temperature is 162 F with a standard deviation of 10 F. Assuming that the temperature of the coffee is approximately normally distributed. Find a 95% confidence interval for the mean temperature.Group of answer choices(172.6, 182.9)(171.5, 181.3)(144.7, 156.5)(168.9, 172.5)(156.7, 167.3)

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

You would like to construct a 99% confidence interval to estimate the population mean price of milk (per gallon) in your city. You select a random sample of prices from different stores. The sample has a mean of 3.49 dollars and a standard deviation of 0.28 dollars.(a) What is the best point estimate, based on the sample, to use for the population mean?dollars(b) For each of the following sampling scenarios, determine which distribution should be used to calculate the critical value for the 99% confidence interval for the population mean.(In the table, Z refers to a standard normal distribution, and t refers to a t distribution.)Sampling scenario Z t Could use either Z or t UnclearThe sample has size 19, and it is from a normally distributed population with a known standard deviation of 0.31. Zt The sample has size 14, and it is from a normally distributed population with an unknown standard deviation. Zt The sample has size 80, and it is from a non-normally distributed population. Zt CheckSave For LaterSubmit Assignment

(c) If appropriate, construct an 80% confidence interval for the mean calorie count for all 12-ounce milkshakes sold at fast-food restaurants. Round the answers to at least two decimal places.An 80% confidence interval for the mean calorie count for all 12-ounce milkshakes sold at fast-food restaurants is <<μ.