Question

Calcium is an essential nutrient for strong bones and for controlling blood pressure and heart beat. Because most of the body’s calcium is stored in bones and teeth, the body withdraws the calcium it needs from the bones. Over time, if more calcium is taken out of the bones than is put in, the result may be thin, weak bones. This is especially important for women who are often recommended a calcium supplement. A consumer group activist assumes that calcium content in two popular supplements are normally distributed with the same unknown population variance, and uses the following information obtained under independent sampling: Supplement 1 Supplement 2 x−1=1,000 mg x−2=1,016 mg s1= 23 mg s2 = 24 mg n1 = 12 n2 = 15 Let μ1 and μ2 denote the corresponding population means. Can we conclude that the average calcium content of the two supplements differs at the 95% confidence level? Multiple Choice No, because the 95% confidence interval contains the hypothesized value of zero. Yes, because the 95% confidence interval contains the hypothesized value of zero. No, because the 95% confidence interval does not contain the hypothesized value of zero. Yes, because the 95% confidence interval does not contain the hypothesized value of zero.

Answer 1

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) ± $t_{n_1+n_2-2;1-\alpha /2}$ * $Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }$]

$t_{n_1+n_2-2;1-/2}= t_{25;0.975}= 2.060$

$Sa= \sqrt{\frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2} }= \sqrt{\frac{11*529+14*576}{12+15-2} } = 23.57$

[(1000-1016)±2.060*23.57*$\sqrt{\frac{1}{12} +\frac{1}{15} }$]

[-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!