Question

1. A local microbrewery claims that its newest beer has at least than 10% alcohol by volume (ABV). To test this claim, a group of drinkers measure the ABV from 25 of their bottled beers. They found that the average ABV is 9.83% with a standard deviation of .50% ABV. Can the drinkers’ measurements support the brewer’s claim, if the test is carried out at the 1% level of significance? What if the test uses a significance level of 10%?

Answer 1

Answer)

Null hypothesis Ho : u>=10%

Alternate hypothesis Ha : u <10%

Given mean = 9.83%

S.d = 0.5%

As the population s.d is unknown here we can use t distribution to conduct the test

Test statistics t = (sample mean - Claimed mean)/(s.d/√n)

test statistics t = (9.83 - 10)/(0.5/√25) = -1.7

Degrees of freedom is = n-1 = 25-1 = 24

For 24 degrees of freedom and -1.7 test statistics

P-Value from t distribution is = 0.051028

We reject null hypothesis when P-Value is less than the given significance level

If significance level is 1% (0.01

Then p-value is greater than 0.01

And we fail to reject the null

So we have enough evidence to conclude that mean is >= 10

If significance level is 0.1 (10%)

Then p-value is less than it

So we reject the null hypothesis

We do not have enough evidence to conclude that mean is >= 10