Question

A baby food company claims that on the average the product contains 6.0 grams of protein per bottle. A random sample of 14 bottles of the product was taken and the protein measurements are as follows. 5.1 4.9 6.0 6.4 5.7 4.8 6.2 6.0 5.9 5.6 5.5 5.8 5.3

(i) Find the sample mean x¯ and the sample standard deviation s.

(ii) Assume that the protein content per bottle obeys an N (µ, σ2) distribution with σ unknown. Give the study’s objective in terms of H0 and HA and then perform a test at the α = 0.05 level.

(iii) In the foregoing analysis (you just performed) you might reject H0 even when the mean protein content is exactly 6.0 gram per bottle as claimed. What is the name for this kind of mistake? What measure can be taken to control the probability of this kind of mistake?

Answer 1

Solution-1:

sample mean =xbar=sum of x/total x=73.2/13=5.6307

  

X	xbar	x-xbar	(x-xbar)^2
5.1	5.6308	-0.5308	0.281749
4.9	5.6308	-0.7308	0.534069
6	5.6308	0.3692	0.136309
6.4	5.6308	0.7692	0.591669
5.7	5.6308	0.0692	0.004789
4.8	5.6308	-0.8308	0.690229
6.2	5.6308	0.5692	0.323989
6	5.6308	0.3692	0.136309
5.9	5.6308	0.2692	0.072469
5.6	5.6308	-0.0308	0.000949
5.5	5.6308	-0.1308	0.017109
5.8	5.6308	0.1692	0.028629
5.3	5.6308	-0.3308	0.109429
		total	2.927692

Sample sd=s=

s=sqrt(2.927692/13-1)

=0.493938

Solutioniii:

Ho:

Ha:

alpha=0.05

perfom t test for mean

t=xbar-mu/s/sqrt(n)

=(5.6308-6)/(0.493938/qrt(13))

t= -2.695013

df=n-1=13-1=12

p value in excel for the t and df is =T.DIST.2T(2.695013,12)

=0.01949

p<0.05

Reject Ho

Accept Ha

Conclusion:

There is no sufficient evidence at 5% level of significance to support the claim that he average the product contains 6.0 grams of protein per bottle

Solutioniii

name for this kind of mistake is type 1 error denoted by alpha

The chance of making a type 1 error are reduced by increasing the level of confidence