Question

Supposed we are interested in understanding the mean number of M&Ms in a small bag of M&M. (Assume the number of M&Ms in a bag follows a normal distribution with unknown mean). You collected a sample of 16 small bags of M&M and obtain the number of counts for each package. We obtained the following sample statistics, mean number of M&M per bag is 25, and standard variation is 4. a) Construct a 90% confidence interval for the true mean number of M&M in a bag. (10pts) b) Conclude your confidence interval. (5 pts)

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 25

Population standard deviation =   2 = 4

=   4 = 2

Sample size = n = 16

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

Margin of error = E = Z/2 * ( /n)

= 1.645 * ( 2 /  16 )

= 0.8225

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

- E < < + E

25 - 0.8225 <   < 25 + 0.8225  

24.1775 <   < 25.8225

( 24.1775 , 25.8225 )

At 90% confidence interval estimate of the population mean is : - ( 24.1775 , 25.8225 )