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Sagot :
Answer:
The pvalue of the test is 0.0178 < 0.1, which means that there is evidence at the .10 level to suggest that the mean IQ score for Lower Arlington residents differs from the national average
Step-by-step explanation:
Previous analyses have determined that national average adult IQ score is 100.
This means that the null hypothesis is: [tex]H_0: \mu = 100[/tex]
Is there any evidence at the .10 level to suggest that the mean IQ score for Lower Arlington residents differs from the national average.
This means that the alternate hypothesis that will be tested is:
[tex]H_a: \mu \neq 100[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
100 is tested at the null hypothesis:
This means that [tex]\mu = 100[/tex]
Standard deviation of 15.
This means that [tex]\sigma = 15[/tex]
A random sample of 35 adults from Lower Arlington, yields an average score of 106.
This means that [tex]n = 35, X = 106[/tex]
Value of the test-statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{106 - 100}{\frac{15}{\sqrt{35}}}[/tex]
[tex]z = 2.37[/tex]
Pvalue of the test:
Since z is positive, and we are testing if the mean is different from a value, the pvalue of the test is 2 multiplied by 1 subtracted by the pvalue of z = 2.37.
Looking at the z-table, z = 2.37 has a pvalue of 0.9911
1 - 0.9911 = 0.0089
2*0.0089 = 0.0178
Is there any evidence at the .10 level to suggest that the mean IQ score for Lower Arlington residents differs from the national average
The pvalue of the test is 0.0178 < 0.1, which means that there is evidence at the .10 level to suggest that the mean IQ score for Lower Arlington residents differs from the national average
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