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.

I would like to know how to Calculate this in Excel.

Calculate the critical value, the test statistic, and p-value.

Make a decision about the null hypothesis and explain your reasoning, then make a conclusion about the claim in nontechnical terms.

Write the hypotheses in symbolic form, determine if the test is right-tailed, left-tailed, or two tailed and explain why.

Low Lead level High Lead Level
n1 78 n2 21
x-bar1 92.88 x-bar2 86.9
s1 15.34 s2 8.99

Solutions

Expert Solution

H0: or mean IQ score of people with low lead levels is equal to the mean IQ score of people with high lead levels.

H1: or mean IQ score of people with low lead levels is higher than the mean IQ score of people with high lead levels.

test statistic:

t=~tn1+n2-2 (putting values from question we get)

t=

t=

t=

t=2.2825~t97,0.05 (calculated)

t97,0.05 =1.661 (tabulated)

p-value is 0.012322<0.05

since, tcal > ttab and also  p-value is 0.012322<0.05 therefore,we reject the null hypothesis.

hence,H1: or mean IQ score of people with low lead levels is higher than the mean IQ score of people with high lead levels is accepted.

please rate my answer and comment for doubts.


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