College

It is known from past years that the average IQ of Commack elementary students is at least 110. The principal thinks that this is an overestimate and that the average IQ of students here is less than 110. The principal administers an IQ test to 20 randomly selected students. Among the sampled students, the average IQ is 104 with a standard deviation of 10. Assume a significance level of 0.05 and conduct a hypothesis test.

Hypothesis Test:
- Null Hypothesis ([tex]H_0[/tex]): The average IQ of Commack elementary students is 110 or more. ([tex]\mu \geq 110[/tex])
- Alternative Hypothesis ([tex]H_a[/tex]): The average IQ of Commack elementary students is less than 110. ([tex]\mu < 110[/tex])

Given Data:
- Sample size ([tex]n[/tex]): 20
- Sample mean ([tex]\bar{x}[/tex]): 104
- Standard deviation ([tex]s[/tex]): 10
- Significance level ([tex]\alpha[/tex]): 0.05

Answer :

Answer:

The average IQ of Commack elementary students is less than 10

Step-by-step explanation:

Null hypothesis: The average IQ of Commack elementary students is 10

Alternate hypothesis: The average IQ of Commack elementary students is less than 10

Test statistic (z) = (sample mean - population mean) ÷ sd/√n

sample mean = 104, population mean = 110, sd = 10, n = 20

z = (104 - 110) ÷ 10/√20 = -6 ÷ 2.24 = -2.68

The test is a one tailed test. The critical value using a 0.05 significance level is 1.645

Conclusion on the null hypothesis: The test statistic (-2.68) is less than the critical value (1.645), reject the null hypothesis

Therefore, the average IQ of Commack elementary students is less than 10.

This result does not support the original claim from past years that the average IQ of Commack elementary students is at least 10.

Final answer:

To conduct a hypothesis test, set up the null and alternative hypotheses. Calculate the test statistic using the t-statistic formula. Compare the calculated t-statistic to the critical t-value at the given significance level to make a conclusion.

Explanation:

To conduct a hypothesis test, we need to set up the null and alternative hypotheses. In this case, since the principal thinks that the average IQ of students at Commack elementary is less than 110, the null hypothesis (H₀) is that the average IQ is equal to or greater than 110, and the alternative hypothesis (H₁) is that the average IQ is less than 110. The significance level (α) is given as 0.05.

Next, we calculate the test statistic. The test statistic for this hypothesis test is the t-statistic, which is calculated as:

t = (sample mean - population mean) / (sample standard deviation / √(sample size)

In this case, the sample mean is 104, the population mean is 110 (from the past years' data), the sample standard deviation is 10, and the sample size is 20. Plugging these values into the formula, we get:

t = (104 - 110) / (10 / √(20)) ≈ -2.529

Finally, we compare the calculated t-statistic to the critical t-value at the given significance level. In this case, since we are testing whether the average IQ is less than 110, it is a left-tailed test. Looking up the critical t-value for α = 0.05 and degrees of freedom (df) = n - 1 = 20 - 1 = 19, we find it to be -1.729. Since the calculated t-statistic (-2.529) is less than the critical t-value (-1.729), we reject the null hypothesis. This means that there is sufficient evidence to conclude that the average IQ of Commack elementary students is less than 110.

Learn more about Hypothesis testing here:

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