Afood snack manufacturer samples 15 bags of pretzels off the assembly line and weighed their contents. if the sample mean is 10.2 and the sample standard deviation is 0.25, find the 95% confidence interval of the true
a. (10.06, 10.34)b. (10.07, 10.33)d. (10.14, 10.26)

the correct option is A

Step-by-step explanation:

We want to find 95% confidence interval for the mean of the weight of pretzels.

Number of samples. n = 15 bags

Mean, u = 10.2

Standard deviation, s = 0.25

We will use the t- test

Degree of freedom = n - 1 = 15 - 1= 14

Alpha, a = (1-confidence interval )/2

a = (1-0.95)/2 = 0.025

Looking at the t-distribution table, the corresponding z value is 2.131

Confidence interval = z × standard deviation/√n

Confidence interval = 2.131 × 0.25/√15

Confidence interval = 0.13755545851

Approximately 0.138

At 95% confidence interval,

The lower end is 10.2 - 0.138 = 10.062 Approximately 10.06

The upper end is 10.2 - 0.138 = 10.338. Approximately 10.34

idk

step-by-step explanation:

use that smart brain

this is false ^.^ lol