A factory manager collected a sample of ten (10) packets of 3-in-1 coffee and the weights (in grams) of each packet are recorded as follow:

25 23 26 28 24 21 26 24 25 27.
Calculate
(a) the mean weight of a packet of 3-in-1 coffee.
(b) the sample standard deviation of the data collected.
(c) find a 90% confidence interval for the weights of the 3-in-1 coffee.

a)Mean =( sum of the observations)/no of observations = ( 46+49+50+53+51)/10 = 249/10= 24 .9   

b) Variance ={(25-24.9)^2. + (21-24.9)^2+ (23-24.9)^2+ (26-24.9)^2+ (24-24.9)^2+(28-24.9)^2+( 25-24.9)^2+(24-24.9)^2+(24-24.9)^2 }/9 ( for unbaised estimated variance)= 4.1. Sd = 2.025

c) here we assume parent population normal distribution

But sampling distribution is a t distribution with (n-1) degrees of freedom

t distribution to be used ( for small sample) n= 10

X is distributed t distribution with (10-1)= 9 degrees of freedom ( for 2 sided tail)

90.% C I is ( mean +/- (t ) sd)/?n) ,t = 1.833 ( for 9 d f)

Where t = 10% significance level with 9 degrees of freedom, from the t distribution table  

90 % Confidence Interval (24 .9 +/- (1.833)( 2.025)/3)  

= ( 24.9 +/- 1.237) = ( 23.663. , 26.137)