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)

A random sample of 13 brands of caffeinated coffee resulted in the following 95% confidence interval for 𝜇, the average caffeine content in a cup of caffeinated coffee: (85.12, 154.08).  Interpret this interval.

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.

(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.

A representative of a consumer advocacy group randomly selects 50 bags of a certain brand of chips and determines the net weight of each. He finds the mean of these selected bags to be 395 grams and the standard deviation to be 6.8 grams. Calculate a 90% confidence interval for the true mean of weight of chips. What is the lower limit of the interval?

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