A quality control specialist for a restaurant chain takes a random sample of size 14 to check the amount of soda served in the 16 oz. serving size. The sample mean is 13.60 with a sample standard deviation of 1.51. Assume the underlying population is normally distributed. Find the 95% confidence interval for the true population mean for the amount of soda served.

The 95% confidence interval = (12.79, 14.41)

Step-by-step explanation:

The confidence interval expresses that the true mean lies within the calculated range with a particular level of confidence.

To obtain the estimate of the 95% confidence interval of the population, we'll need the standard error

Upper limit of the confidence interval = (sample mean) + (standard error)

Lower limit of the confidence interval = (Sample mean) - (standard error)

Standard error of the sample mean = (critical value) × (standard deviation of the sample mean)

(Standard deviation of the sample mean) = (standard deviation)/√n

where n = sample size = 14

(Standard deviation of the sample mean) = (1.55/√n) = (1.55/√14) = 0.4143

Critical value for 95% confidence interval = z = 1.96 (z-score from the tables)

Standard error of the sample mean = 1.96 × 0.4143 = 0.81

Upper limit of the confidence interval = (sample mean) + (standard error) = 13.60 + 0.81 = 14.41

Lower limit of the confidence interval = (Sample mean) - (standard error) = 13.60 - 0.81 = 12.79

The 95% confidence interval = (12.79, 14.41)

Hope this Helps!!!

4400

step-by-step explanation:

answer:

step-by-step explanation:

here, the total dishes = 10

dishes having meat = 3 ( chicken, meatloaf and hamburger )

thus, the probability of having meat on the chosen dishes = number of dishes having meat / total dishes

=

therefore, the probability of not having meat = 1- the probability of having meat on the chosen dishes

=

=

=

⇒ option b is correct.