Mathematics, 01.07.2020 16:01, breadwonders3738

The Bayley Scales of Infant Development yield scores on two indices--the Psychomotor Development Index (PDI) and the Mental Development Index (MDI)--which can be use to assess a child's level of functioning in each of these areas at approximately one year of age. Among normal healthy infants, both indices have a mean value of 100. As part of a study assessing the development and neurologic status of children who have undergone reparative heart surgery during the first three months of life, the Bayley Scales were administered to a sample of one-year-old infants with congenital heart disease. The data contained in the data set heart. PDI scores are saved under the variable name pdi while MDI scores are saved under mdi. Use the treatment=1 group a. At the 0.05 level of significance, test the null hypothesis that the mean PDI score for children born with congenital heart disease who undergo reparative heart surgery during the first three months of life is equal to 100, the mean score for healthy children. Use a two-sided test. What is the p-value? What do you conclude?
b. Conduct the analogous test of hypothesis for the mean MDI score. What do you conclude?
c. Construct 95% confidence intervals for the true mean PDI score and the true mean MDI score for this population of children with congenital heart disease. Does either of these intervals contain 100? Would you have expected that they would?

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answered: cuppykittyy

Step-by-step explanation:

Hello!

The Psychomotor Development Index (PDI) has an average value of μ= 100

The Mental Development Index (MDI) has an average value of μ= 100

At an approximate age of 1 year of normal healthy infants.

Using the group data set heart = 1 for all calculations (see attachment for complete table), you can define two variables of interest and obtain the descriptive statistics:

X₁: PDI of an infant with congenital heart disease who had to undergo reparative heart surgery during the first three months of life.

n₁= 69

X[bar]₁= 97.61

S₁= 14.73

X₂: MDI of an infant with congenital heart disease who had to undergo reparative heart surgery during the first three months of life.

n₂= 69

X[bar]₂= 106.33

S₂= 14.67

You have to test the hypothesis that the average PDI for kids with congenital heart disease who have to undergo reparative heart surgery during the first three months of life is equal to 100.

H₀: μ₁ = 100

H₁: μ₁ ≠ 100

α: 0.05

$Z= \frac{X[bar]_1-Mu_1}{\frac{S_1}{\sqrt{n_1} } }$ ≈N(0;1)

$Z_{H_0}= \frac{97.61-100}{\frac{14.73}{\sqrt{69} } } = -1.347$

p-value: 0.17798

Using this approach the decision rule is:

If p-value ≤ α, reject the null hypothesis.If p-value > α, do not reject the null hypothesis.

The p-value is greater than the level of significance, the decision is to not reject the null hypothesis. Then the average PDI of the kids with congenital heart disease who have to undergo reparative heart surgery during the first three months of life is equal to 100.

H₀: μ₂ = 100

H₁: μ₂ ≠ 100

α: 0.05

$Z= \frac{X[bar]_2-Mu_2}{\frac{S_2}{\sqrt{n_2} } }$ ≈N(0;1)

$Z_{H_0}= \frac{106.33-100}{\frac{14.67}{\sqrt{69} } } = 3.584$

p-value: 0.000338

Using this approach the decision rule is:

If p-value ≤ α, reject the null hypothesis.If p-value > α, do not reject the null hypothesis.

The p-value is less than the level of significance, the decision is to reject the null hypothesis. Then the average MDI of the kids with congenital heart disease who have to undergo reparative heart surgery during the first three months of life is different from 100.

$Z_{1-\alpha /2}= Z_{0.975}= 1.96$

95% CI for PDI

X[bar]₁ ± $Z_{1-\alpha /2}$ * $\frac{S_1}{\sqrt{n_1} }$

97.61 ± 1.96 * $\frac{14.73}{\sqrt{69} }$

[94.134; 101.086]

95% CI for MDI

X[bar]₂ ± $Z_{1-\alpha /2}$ * $\frac{S_2}{\sqrt{n_2} }$

106.33 ± 1.96 * $\frac{14.67}{\sqrt{69} }$

[102.869; 109.791]

The CI for the mean PDI of the kids with congenital heart disease who have to undergo reparative heart surgery during the first three months of life contains 100, this is to be expected since the null hypothesis in the hypothesis test made at complementary confidence level, was not rejected.

I hope this helps!