Question

At Ajax Spring Water, a half-liter bottle of soft drink is supposed to contain a mean of 520 ml. The filling process follows a normal distribution with a known process standard deviation of 4 ml.

(a)	The normal distribution should be used for the sample mean because (You may select more than one answer. Click the box with a check mark for the correct answer and click to empty the box for the wrong answer.)
	the sample population has a large mean. the population distribution is known to be normal. the population standard deviation is known. the standard deviation is very small.

(b)	Set up hypotheses and a two-tailed decision rule for the correct mean using the 5 percent level of significance.
	The hypothesis for a two-tailed decision is

a.	H0: μ ≠520, H1: μ = 520, reject if z < −1.96 or z > 1.96
b.	H0: μ ≠520, H1: μ = 520, reject if z > 1.96 or z < −1.96
c.	H0: μ = 520, H1: μ ≠ 520, reject if z > 1.96 or z < −1.96
d.	H0: μ =520, H1: μ ≠ 520, reject if z > −1.96 or z < 1.96

	a b c d

(c)	If a sample of 16 bottles shows a mean fill of 515 ml, does this contradict the hypothesis that the true mean is 520 ml?
	Yes No

Answer 1

a) Since the population standard deviation is known . if population standard deviation is known and samples are comming from normal population we use normal distribution instead of t-distribution to test the sample mean.

Hence corrective choices are i) the population distribution known to be normal.

ii) the population standard deviation is known.

b) for two tailed test we test the hypothesis

H₀: =₀ against H₁:#₀

and decision criteria at 5% level of significance is

reject H0 if z > 1.96 if z takes positive value and Z < -1.96 if z takes negative value

where z is test-statistic based on normal distribution

Hence the correct choice is

The hypothesis for a two-tailed decision is

c: Ho: = 520 , H1: # 520 reject if z > 1.96 and z < -1.96

c) from the information n=16 , xbar=515 and =4

we want to test the hypothesis

Ho: = 520 , H1: # 520

Under Ho the test statistic is

z= (xbar- ) / ( /sqrt(n)) ~ N(0,1)

z= ( 515-520)/(4/4)

= -5

Table value z _alpha/2 = z _0.025 = 1.96

Since Z < -1.96 hence  we reject the null hypothesis.

Conclusion : There is contradiction the hypothesis that the true mean is 520 ml.

Corrective choice : Yes