Answer:
No, because the 95% confidence interval contains the hypothesized value of zero.
Step-by-step explanation:
Hello!
You have the information regarding two calcium supplements.
X₁: Calcium content of supplement 1
n₁= 12
X[bar]₁= 1000mg
S₁= 23 mg
X₂: Calcium content of supplement 2
n₂= 15
X[bar]₂= 1016mg
S₂= 24mg
It is known that X₁~N(μ₁; σ²₁), X₂~N(μ₂;δ²₂) and σ²₁=δ²₂=?
The claim is that both supplements have the same average calcium content:
H₀: μ₁ - μ₂ = 0
H₁: μ₁ - μ₂ ≠ 0
α: 0.05
The confidence level and significance level are to be complementary, so if 1 - α: 0.95 then α:0.05
since these are two independent samples from normal populations and the population variances are equal, you have to use a pooled variance t-test to construct the interval:
[(X[bar]₁-X[bar]₂) ± *
]
[(1000-1016)±2.060*23.57*]
[-34.80;2.80] mg
The 95% CI contains the value under the null hypothesis: "zero", so the decision is to not reject the null hypothesis. Then using a 5% significance level you can conclude that there is no difference between the average calcium content of supplements 1 and 2.
I hope it helps!