Answer:
h0:u ≥ 21.1
h1: u < 21.1
t statistics = -2.632
pvalu = 0.007
there is sufficient evidence to support claim
Explanation:
null hypothesis
h0 = u≥21.1
alternative = u < 21.1
to get test statistics
barx - u / s/√n
= 19.4 - 21.1 / 3.23/√25
= -2.632
we calculate for the p value given this test statistics and the degree of freedom = 25-1 = 24
lest tailed p value = 0.0073 this is approximately 0.007
p value is less than the level of significance so we reject h0
we conclude that we have enough evidence to support h1, mean tar conent is less than 21.1