A researcher is testing a null hypothesis that states: H: ì = 50. A sample of 25 scores is selected and the mean is M = 55.Assuming that the sample variance is s2 = 100, compute the estimated standard error and the t statistic. Is this sample sufficient to reject the null hypothesis using a two-tailed test with á = .05? Assuming that the sample variance is s2 = 400, compute the estimated standard error and the t statistic. Is this sample sufficient to reject the null hypothesis using a two-tailed test with á = .05? Explain how increasing variance affects the standard error and the likelihood of rejecting the null hypothesis.
df = 25-1 = 24
For P = .05, with 24 df, you need to have difference between means ± 2.064(SEdiff)
t = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1) or to make it easier, you can use n
SD = √(variance)
I'll leave the calculation and explanation to you.
To compute the estimated standard error and the t statistic, we need to use the formula for the standard error of the mean (SE) and the formula for the t statistic.
The formula for the standard error of the mean is:
SE = sqrt(s^2 / n),
where s^2 is the sample variance and n is the sample size.
In the first scenario, where s^2 = 100 and n = 25, we can calculate the estimated standard error as follows:
SE = sqrt(100 / 25) = sqrt(4) = 2.
Now, let's calculate the t statistic. The formula for the t statistic is:
t = (M - μ) / (SE),
where M is the sample mean and μ is the null hypothesis mean.
In this case, M = 55 and μ = 50. Plugging in the values, we get:
t = (55 - 50) / 2 = 5 / 2 = 2.5.
To determine whether the sample is sufficient to reject the null hypothesis, we need to compare the calculated t statistic with the critical t value at a chosen significance level (α). In this case, α = 0.05 for a two-tailed test.
The critical t value can be looked up in a t-distribution table or using statistical software. For a two-tailed test with α = 0.05 and 23 degrees of freedom (n - 1 = 25 - 1 = 24), the critical t value is approximately ±2.064.
Since the calculated t statistic (2.5) is greater than the critical t value (2.064), we can reject the null hypothesis. Therefore, the sample is sufficient to reject the null hypothesis with a two-tailed test at α = 0.05.
Let's move on to the second scenario, where s^2 = 400. The estimated standard error can be calculated as follows:
SE = sqrt(400 / 25) = sqrt(16) = 4.
With the new value of s^2, the standard error increases to 4. Now, let's calculate the t statistic:
t = (55 - 50) / 4 = 5 / 4 = 1.25.
Using the same critical t value for a two-tailed test with α = 0.05 and 24 degrees of freedom (n - 1 = 25 - 1 = 24), which is approximately ±2.064, we find that the calculated t statistic (1.25) is smaller than the critical t value (2.064). Therefore, in the second scenario, we fail to reject the null hypothesis.
When the variance increases from 100 to 400, the standard error also increases. As the standard error increases, it implies greater uncertainty in estimating the true population mean. When the standard error is larger, the t statistic becomes smaller because the difference between the sample mean and the null hypothesis mean is relatively smaller when compared to the increased variability. Consequently, the likelihood of rejecting the null hypothesis decreases as the standard error increases.