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

1 answer

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.