To test the hypothesis, we can use the one-sample t-test since we know the sample mean, sample size, population mean, and population standard deviation.
Null hypothesis (H0): The population mean IQ is equal to 100. μ = 100
Alternative hypothesis (H1): The population mean IQ is not equal to 100. μ ≠ 100
Test statistic formula:
t = (x̄ - μ) / (s / sqrt(n))
where:
x̄ = Sample mean (108)
μ = Population mean (100)
s = Standard deviation (10)
n = Sample size (51)
Now let's calculate the test statistic:
t = (108 - 100) / (10 / sqrt(51))
t = 8 / (10 / sqrt(51))
t ≈ 8 / 1.397
t ≈ 5.72
Using the t-distribution table or a statistical software, we can find the critical t-value for a 5% significance level with 50 degrees of freedom. The critical t-value (two-tailed) is approximately ±2.009.
Since the calculated test statistic (5.72) is greater than the critical t-value (2.009), we can reject the null hypothesis.
Therefore, at a 5% significance level, we have enough evidence to conclude that the population mean IQ is significantly different from 100.
The IQ’s of 51 students are tested and it is found that the average IQ is 108 with a standard deviation of 10. Test using a 5% significance level. Calculate the test statistic and outcome of the test
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