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Question
Math
Posted 6 months ago

The IQ scores of a population are normally distributed with a mean of 100 and a standard deviation of 15. A psychologist wishes to test the theory that eating chocolate before sitting an IQ test improves your score. A random sample of 80 people are selected and they are each give an identical bar of chocolate to eat before taking an IQ test.
a Find, at the 2.5%2.5 \% level, the critical region for this test, stating your hypotheses clearly.
The mean score on the test for the sample of 80 people was 102.5 .
b Comment on this observation in light of the critical region.
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Answer from Sia
Posted 6 months ago
Solution by Steps
step 1
Define the null hypothesis H0H_0 and the alternative hypothesis H1H_1
$H_0$: Eating chocolate does not improve IQ scores (mean = 100).
$H_1$: Eating chocolate improves IQ scores (mean > 100).
step 2
Determine the critical value for a one-tailed test at the 2.5%2.5\% significance level
For a one-tailed test at the $2.5\%$ significance level, the critical value $z$ is approximately $1.96$.
step 3
Calculate the standard error of the mean (SEM)
SEM = $\frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{80}} \approx 1.677$
step 4
Calculate the test statistic zz using the sample mean xˉ=102.5\bar{x} = 102.5
$z = \frac{\bar{x} - \mu}{SEM} = \frac{102.5 - 100}{1.677} \approx 1.49$
step 5
Compare the test statistic to the critical value
Since $1.49 < 1.96$, we fail to reject the null hypothesis $H_0$.
Answer
The critical region for this test is z>1.96z > 1.96. Since the test statistic z=1.49z = 1.49 is less than 1.961.96, we fail to reject the null hypothesis.
Key Concept
Hypothesis Testing
Explanation
In hypothesis testing, we compare the test statistic to a critical value to determine whether to reject the null hypothesis. In this case, the test statistic did not exceed the critical value, so we do not have enough evidence to support the claim that eating chocolate improves IQ scores.

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