Question

= 3. The past five days of plant operation have resulted in the following yields 91.6, 88.75, 90.8, 89.94, and 91.3. The yield of a chemical process is being studied. From previous experience, yield is known to be normally distributed and Find a 95% two-sided confidence interval on the true mean yield. (a) Calculate the sample mean. Round your answer to 2 decimal places.

Answer 1

Solution :

Given that,

a)

sample mean = $\bar x$ = sum ( X ) / n = ( 91.6 + 88.75 + 90.8 + 89.94 + 91.3 ) / 5 = 90.48

Population standard deviation = $\sigma$ = 3

Sample size = n = 5

b)

At 95% confidence level the z is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z_$\alpha$/2 = Z_0.025 = 1.96

Margin of error = E = Z$\alpha$_/2* ($\sigma$ /$\sqrt$n)

= 1.96 * ( 3/ $\sqrt$ 5 )

= 1.34

At 95% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

90.48 - 1.34 < $\mu$ < 90.48 + 1.34

89.14 < $\mu$ < 91.82

(89.14 , 91.82)