Question
In: Statistics and Probability
A researcher conducted a test to learn the effect of lead levels in human bodies. He...
A researcher conducted a test to learn the effect of lead levels in human bodies. He collected the IQ scores for a random sample of subjects with low lead levels in their blood and another random sample of subjects with high lead levels in their blood. The summary of finding is listed below. Use a 0.05 significance level to test the claim that the mean IQ score of people with low lead levels is higher than the mean IQ score of people with high lead levels.
We do not know the values of the population standard deviations.
| Low Lead Level | High Lead Level | |||
| n1 | 78 |
|
21 | |
|
92.88 | x2bar | 86.9 | |
| s1 | 15.34 | s2 | 8.99 | |
1. Write the hypotheses in symbolic form, determine if the test is right-tailed, left-tailed, or two tailed and explain why.
2. Calculate the critical value, the test statistic, and p-value. Show calculations below.
3. Make a decision about the null hypothesis and explain your reasoning, then make a conclusion about the claim in nontechnical terms.
Solutions
orchestra answered 2 years agoRelated Solutions
A researcher conducted a test to learn the effect of lead levels in human bodies. He...
A researcher conducted a test to learn the effect of lead levels
in human bodies. He collected the IQ scores for a random sample of
subjects with low lead levels in their blood and another random
sample of subjects with high lead levels in their blood. The
summary of finding is listed below. Use a 0.05 significance level
to test the claim that the mean IQ score of people with low lead
levels is higher than the mean IQ score...
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An experiment was conducted to test the effect of a new drug on a viral infection....
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a viral infection. After the infection was induced in 100 mice, the
mice were randomly split into two groups of 50. The control group
received no treatment for the infection, while the other group
received the drug. After a 30-day period, the proportions of
survivors, p̂1 and p̂2, in the two groups were found to be 0.38 and
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An experiment was conducted to test the effect of a new drug on a viral infection....
An experiment was conducted to test the effect of a new drug on
a viral infection. After the infection was induced in 100 mice, the
mice were randomly split into two groups of 50. The first group,
the control group, received no treatment for the infection, and the
second group received the drug. After a 30-day period, the
proportions of survivors, p̂1 and p̂2, in the two groups were found
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An experiment was conducted to test the effect of a new drug on a viral infection....
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a viral infection. After the infection was induced in 100 mice, the
mice were randomly split into two groups of 50. The first group,
the control group, received no treatment for the
infection, and the second group received the drug. After a 30-day
period, the proportions of survivors, p?1 and
p?2, in the two groups were found to be 0.32
and 0.60, respectively.
(a) Is there sufficient...
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