the IQ’s of 20 students are tested and it is found that the average IQ is 117 with a variance of 121. Test H0:µ = 110 vs Ha:µ ≠ 110 using a 1% significance level. Calculate the test statistic and outcome of the test.
To test the hypothesis H0: µ = 110 vs Ha: µ ≠110, we can use a t-test since the population variance is unknown.
Given:
Sample mean (x̄) = 117
Population mean (µ) = 110
Sample size (n) = 20
Variance (σ^2) = 121
First, we need to calculate the test statistic using the formula:
t = (x̄ - µ) / (s / √n)
where
s = √(σ^2) is the standard deviation
s = √121 = 11
t = (117 - 110) / (11 / √20)
t = 7 / (11 / √20)
t ≈ 4.53
Next, we need to compare the test statistic with the critical value at a 1% significance level. Since the alternative hypothesis is two-sided (µ ≠ 110), we will use a two-tailed test.
Looking up the critical value for a two-tailed test with a 1% significance level and 19 degrees of freedom (n - 1 = 20 - 1 = 19), we find t_critical ≈ ±2.86.
Since the absolute value of the calculated test statistic (|4.53|) is greater than the critical value (2.86), we reject the null hypothesis.
Therefore, the outcome of the test is that there is sufficient evidence to suggest that the true population mean IQ is not equal to 110.
To test the hypothesis H0: µ = 110 vs Ha: µ ≠ 110, we can use a t-test since the population variance is unknown.
Given:
Sample size (n) = 20
Sample mean (x̄) = 117
Sample variance (s²) = 121
Population mean (µ₀) = 110
Step 1: Set up the hypothesis
Null hypothesis (H0): µ = µ₀ (Population mean is 110)
Alternative hypothesis (Ha): µ ≠ µ₀ (Population mean is not equal to 110)
Step 2: Determine the significance level and choose the appropriate test statistic
The significance level is given as 1%, which corresponds to a 2-tailed test. Therefore, the critical value is ±2.845 (obtained from t-distribution table with (n-1) degrees of freedom, where n = 20 and α = 0.01).
Step 3: Calculate the test statistic
The test statistic (t) can be calculated using the formula:
t = (x̄ - µ₀) / (s / √n)
Substituting the given values:
t = (117 - 110) / (√(121 / 20))
t = 7 / (√(6.05))
t ≈ 7 / 2.462
t ≈ 2.843
Step 4: Make a decision
Since the calculated test statistic (t = 2.843) falls within the range of ±2.845, we fail to reject the null hypothesis.
Outcome of the test:
Based on the given data and calculations, we do not have sufficient evidence to reject the null hypothesis (H0: µ = 110) at the 1% significance level.