Question

Guinness Ghana Breweries limited, producers of Malta Guinness indicates on the label that the bottle contains 330ml of malt drink. A sample of 75 bottles is selected hourly and the contents weighed. Last hour a sample of 75 bottles had a mean weight of 338ml with a standard deviation of 4ml. At the 5% significance level;                                                         
i. State the null hypothesis and the alternative hypothesis.
ii. State the decision rule
iii. At the .05 significance level is the process out of control?

Answer 1

i) Claim: The average of the bottle contains 330ml of malt drink.
From the above claim the null hypothesis(H₀) and the alternative hypothesis(H₁) are as follows:
H₀: µ = 330ml
H₁: µ ≠ 330ml

ii) Decision rule:
1) If p-value < level of significance (α) then we reject null hypothesis
2) If p-value > level of significance (α) then we fail to reject null hypothesis.
(Note: Here I am using p-value method)

iii) Let's write the given information.
= sample mean = 338ml,
s = sample standard deviation = 4ml
μ = 330ml (from the null hypothesis).
n = sample size = 75.
Test Statistic:
Here population standard deviation(σ) is not given and we use sample standard deviation(s) instead of population standard deviation. The sample size(n = ) is small (< 30) and we need to assuming that the sample comes from the normal population, so we can used one sample t test for testing the above hypothesis.
The formula of t-test statistic is as follows:
..(1)
Plugging the above values in equation(1), we get

The test statistic = 17.321.
p-Value:
We find the value of the test statistic is = t = 17.321
For two-tailed t-test, p-value = 2*P(T > |t|)
Here we need to use technology as "Excel" to find the p-value.
The general command to find p-value for one-tailed t-test is "=TDIST(|t|,df,2)" (Here "2" for two tailed test)
df = degrees of freedom = n - 1 = 75 - 1 = 74
so p-value = "=TDIST(17.321,74,2)" ≈ 0
p-value = 0.
Since p value = 0 < 0.05 so we used first rule.
That is we reject the null hypothesis.
Decision: Reject the null hypothesis(H₀).
There are sufficient evidence to conclude that the true average of the bottle contains is different than the 330ml of malt drink.
Interpretation: At the .05 significance level the process is out of control.