a food snack manufacturer samples 9 bags of pretzels off the assembly line and weights their contents. If the sample mean is 14.2 oz. and teh sample devision is 0.70 oz, find the 95% confidense interval of the true mean

Step-by-step explanation:

The formula for calculating the confidence interval is expressed as shown;

CI = xbar ± Z(б/√n)

xbar is the sample mean

Z is the value at 95% confidence interval

б is the standard deviation of the sample

n is the number of samples

Given xbar = 14.2, Z at 95% CI = 1.96, б = 0.70 and n = 9

Substituting this values into the formula;

CI = 14.2 ± 1.96(0.70/√9)

CI = 14.2 ± 1.96(0.70/3)

CI = 14.2 ± 1.96(0.2333)

CI = 14.2 ± 0.4573

CI = (14.2-0.4573, 14.2+0.4573)

CI = (13.7427, 14.6537)

Hence, the 95% confidence interval of the true mean is within the range

13.7≤≤14.7 (to 1 decimal place).

29.5977692753

step-by-step explanation:

i think it will be zero as slope has a formula of difference between initial and final distance divided by itz time interval