Question

Stacy wants to advertise how many chocolate chips (in grams) are in each Big Chip muffin at her bakery. She randomly selects a sample of 19 muffins and finds that the number of chocolate chips per muffin in the sample has a mean of 118.3 and a standard deviation of 25.7. What is the 90% confidence interval for the number of chocolate chips per muffin for Big Chip muffins?

Answer 1

Solution :

Given that,

t_{$\alpha$ /2,df} = 1.734

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 1.734 * (25.7 / $\sqrt$ 19)

Margin of error = E = 10.2

The 90% confidence interval estimate of the population mean is,

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

118.3 - 10.2 < $\mu$ < 118.3 + 10.2

108.1 < $\mu$ < 128.5

(108.1 , 128.5)