Question

the berliner jelly donut company claims that its donuts contain a mean of m grams of jelly. A sample of 100 donuts from the bakery line shows a mean of 78g of jelly with a standard deviation of 12 g. what is a reasonable value for m at a confidence level of 95%?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 78 g

sample standard deviation = s = 12 g

sample size = n = 100

Degrees of freedom = df = n - 1 = 100 - 1 = 99

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t0.025,99 = 1.984

Margin of error = E = t/2,df * (s /n)

= 1.984 * ( 12 / 100)

Margin of error = E = 2.38

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

  ± E  

= 78  ± 2.38

= ( 75.62 g, 80.38 g )