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Furkat [3]
3 years ago
15

Students at a local elementary school were shown a painting and asked which emotion—joy, happiness, love, or anger—they felt by

looking at the painting. The students were classified by their age. The following table summarizes the responses of the students by age-group. One student from the school will be selected at random. What is the probability that the student is in the age-
group of 6 to 8 years given that the selected student responded joy?
Mathematics
1 answer:
Ilya [14]3 years ago
8 0

Answer:

28/89

Step-by-step explanation:

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Answer:

99% CI:

-4.637\leq\mu_v-\mu_o\leq 3.737

Step-by-step explanation:

We have to calculate a 99%CI for the difference of means for the vegan and the omnivore.

First, we have to estimate the standard deviation

s_{Md}=\sqrt{\frac{2MSE}{n_h}}

The MSE can be calculated as

MSE=\frac{(SSE_1+SSE_2)}{df} =\frac{(n_1s_1)^2+(n_2*s_2)^2}{n_1+n_2-2}\\\\MSE=\frac{(85*1.05)^2+(96*1.20)^2}{85+96-2}=\frac{7965.56+13272.04}{179} =118.65

The weighted sample size nh can be calculated as

n_h=\frac{2}{1/n_1+1/n_2}=\frac{2}{1/85+1/96}=\frac{2}{0.0222} =90.16

The standard deviation then becomes

s_{Md}=\sqrt{\frac{2MSE}{n_h}}=\sqrt{\frac{2*118.65}{90.16}}=\sqrt{2.632} =1.623

The z-value for a 99% confidence interval is z=2.58.

The difference between means is

\Delta M=M_v-M_o=5.10-5.55=-0.45

Then the confidence interval can be constructed as

\Delta M-z*s_{Md}\leq\mu_v-\mu_o\leq \Delta M+z*s_{Md}\\\\ -0.45-2.58*1.623\leq\mu_v-\mu_o\leq -0.450+2.58*1.623\\\\-0.450-4.187\leq\mu_v-\mu_o\leq-0.450+4.187\\\\-4.637\leq\mu_v-\mu_o\leq 3.737

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