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
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.
To calculate the test statistic and outcome of the test, we need to perform a hypothesis test.
Step 1: State the Hypotheses:
The null hypothesis (H0): The average IQ of the population is equal to 100.
The alternative hypothesis (Ha): The average IQ of the population is not equal to 100.
Step 2: Set the Significance Level:
Given that the significance level is 5%, we can express it as α = 0.05.
Step 3: Calculate the Test Statistic:
The test statistic for comparing a sample mean to a population mean when the population standard deviation is known is given by:
test statistic (Z) = (sample mean - population mean) / (population standard deviation / square root of sample size)
Here:
sample mean (x̄) = 108
population mean (µ) = 100
population standard deviation (σ) = 10
sample size (n) = 51
Z = (108 - 100) / (10 / √51)
Z = 8 / (10 / 7.141)
Z = 8 / 1.4
Z = 5.71
Step 4: Determine the Outcome of the Test:
Since the alternative hypothesis is that the average IQ is not equal to 100, we will perform a two-tailed test.
Using the Z-table or a statistical software, we can find the critical Z-values for the given significance level. For a 5% significance level (two-tailed test), the critical Z-values are approximately -1.96 and +1.96.
If the calculated test statistic (Z) is outside this range, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.
In this case, the calculated test statistic (Z) is 5.71, which is far greater than +1.96, so we can reject the null hypothesis.
Therefore, based on the given data and using a 5% significance level, we can conclude that there is sufficient evidence to reject the claim that the average IQ of the population is equal to 100.