Question

Suppose parts (a) through (d) below provide results for a study on the role of calcium in reducing the symptoms of PMS. For each of the parts, compute an approximate 95% confidence interval for the difference in mean symptom scores between the placebo and calcium-treated conditions for the symptom listed. In each case, the results given are mean ± standard deviation. Suppose there were 228 participants in the placebo group and 212 in the calcium-treated group. (Round your answers to two decimal places.)


(a) mood swings: placebo = 0.70 ± 0.78; calcium = 0.50 ± 0.53

(b) crying spells: placebo = 0.39 + 0.57; calcium = 0.21 + 0.40

(c) aches and pains: placebo = 0.45 + 0.60; calcium = 0.37 + 0.45

(d) craving sweets or salts: placebo = 0.60 + 0.75; calcium = 0.44 + 0.61

Answer 1

Answer:

Step-by-step explanation:

Hello!

To test if calcium reduces the symptoms of PMS two independent groups of individuals are compared, the first group, control, is treated with the placebo, and the second group is treated with calcium.

The parameter to be estimated is the difference between the mean symptom scores of the placebo and calcium groups, symbolically: μ₁ - μ₂

There is no information about the distribution of both populations X₁~? and X₂~? but since both samples are big enough, n₁= 228 and n₂= 212, you can apply the central limit theorem and approximate the sampling distribution to normal X[bar]₁≈N(μ₁;δ₁²/n) and X[bar]₂≈N(μ₂;δ₂²/n)

The formula for the CI is:

[(X[bar]₁-X[bar]₂) ± $Z_{1-\alpha /2}$ * $\sqrt{\frac{S^2_1}{n_1} +\frac{S^2_2}{n_2} }$]

95% confidence level $Z_{1-\alpha /2}= Z_{0.975}= 1.96$

(a) mood swings: placebo = 0.70 ± 0.78; calcium = 0.50 ± 0.53

X₁: Mood swings score of a participant of the placebo group.

X₂: Mood swings score of a participant of the calcium group.

[(0.70-0.50) ± 1.96 * $\sqrt{\frac{0.78^2}{228} +\frac{0.53^2}{212} }$]

[0.076; 0.324]

(b) crying spells: placebo = 0.39 + 0.57; calcium = 0.21 + 0.40

X₁: Crying spells score of a participant of the placebo group.

X₂: Crying spells score of a participant of the calcium group.

[(0.39-0.21) ± 1.96 * $\sqrt{\frac{0.57^2}{228} +\frac{0.40^2}{212} }$]

[0.088; 0.272]

(c) aches and pains: placebo = 0.45 + 0.60; calcium = 0.37 + 0.45

X₁: Aches and pains score of a participant of the placebo group.

X₂: Aches and pains score of a participant of the calcium group.

[(0.45-0.37) ± 1.96 * $\sqrt{\frac{0.60^2}{228} +\frac{0.45^2}{212} }$]

[-0.019; 0.179]

(d) craving sweets or salts: placebo = 0.60 + 0.75; calcium = 0.44 + 0.61

X₁: Craving for sweets or salts score of a participant of the placebo group.

X₂: Craving for sweets or salts score of a participant of the calcium group.

[(0.60-0.44) ± 1.96 * $\sqrt{\frac{0.75^2}{228} +\frac{0.61^2}{212} }$]

[0.032; 0.287]

I hope this helps!

Answer 2

Using the z-distribution, the 95% confidence intervals are:

a) (0.08, 0.32).

b) (0.09, 0.27).

c) (-0.02, 0.18).

d) (0.03, 0.29).

We have to find the critical value, which is z with a p-value of $\frac{1 + \alpha}{2}$, in which $\alpha$ is the confidence level.

In this problem, $\alpha = 0.95$, thus, z with a p-value of $\frac{1 + 0.95}{2} = 0.975$, which means that it is z = 1.96.

Item a:

The standard errors are:

$s_P = \frac{0.78}{\sqrt{228}} = 0.0517$

$s_C = \frac{0.53}{\sqrt{212}} = 0.0364$

For the distribution of the differences, we have that:

$\overline{x} = \mu_P - \mu_C = 0.7 - 0.5 = 0.2$

$s = \sqrt{s_P^2 + s_C^2} = \sqrt{0.0517^2 + 0.0364^2} = 0.0632$

The interval is:

$\overline{x} \pm zs$

Hence:

$\overline{x} - zs = 0.2 - 1.96(0.0632) = 0.08$

$\overline{x} + zs = 0.2 + 1.96(0.0632) = 0.32$

The interval is (0.08, 0.32).

Item b:

The standard errors are:

$s_P = \frac{0.57}{\sqrt{228}} = 0.03775$

$s_C = \frac{0.4}{\sqrt{212}} = 0.02747$

For the distribution of the differences, we have that:

$\overline{x} = \mu_P - \mu_C = 0.39 - 0.21 = 0.18$

$s = \sqrt{s_P^2 + s_C^2} = \sqrt{0.03775^2 + 0.02747^2} = 0.0467$

Hence:

$\overline{x} - zs = 0.18 - 1.96(0.0467) = 0.09$

$\overline{x} + zs = 0.18 + 1.96(0.0467) = 0.27$

The interval is (0.09, 0.27).

Item c:

The standard errors are:

$s_P = \frac{0.6}{\sqrt{228}} = 0.0397$

$s_C = \frac{0.45}{\sqrt{212}} = 0.0309$

For the distribution of the differences, we have that:

$\overline{x} = \mu_P - \mu_C = 0.45 - 0.37 = 0.08$

$s = \sqrt{s_P^2 + s_C^2} = \sqrt{0.0397^2 + 0.0309^2} = 0.0503$

Hence:

$\overline{x} - zs = 0.08 - 1.96(0.0503) = -0.02$

$\overline{x} + zs = 0.08 + 1.96(0.0503) = 0.18$

The interval is (-0.02, 0.18).

Item d:

The standard errors are:

$s_P = \frac{0.75}{\sqrt{228}} = 0.0497$

$s_C = \frac{0.61}{\sqrt{212}} = 0.0419$

For the distribution of the differences, we have that:

$\overline{x} = \mu_P - \mu_C = 0.60 - 0.44 = 0.16$

$s = \sqrt{s_P^2 + s_C^2} = \sqrt{0.0497^2 + 0.0419^2} = 0.065$

Hence:

$\overline{x} - zs = 0.16 - 1.96(0.065) = 0.03$

$\overline{x} + zs = 0.16 + 1.96(0.065) = 0.29$

The interval is (0.03, 0.29).

A similar problem is given at https://brainly.com/question/15297663