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)

answer: 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 2.262

confidence interval = z × standard deviation/√n

confidence interval = 2.262 × 0.25/√15

confidence interval = 0.14601147215

approximately 0.146

at 95% confidence interval,

the lower end is 10.2 - 0.146 = 10.054. approximately 10.06

the upper end is 10.2 + 0.146 = 10.346. approximately 10.35

answer:

the answer is the second option: 1.23

step-by-step explanation:

to solve this problem we use the logarithmic base change property.

let c, d and h be positive real numbers then:

we also know that:

is also written as .

then, we can write:

then:

so:

the answer is 4 minutes 30 seconds.