Please analyze the following data using the T-test for Unpaired data:
A school district wants to determine if the girls are equal to boys in test scores. They took a random sample of 22 8th grade girls and 24 8th grade boys. The district looked at recent CAT scores and found the mean score for girls to be xbar1 = 25 and the mean score for boys to be xbar2 = 26. The standard deviation for the girls is s1 = 2.2 and for the boys the standard deviation is s2 = 3.4. Is the school district correct in assuming the girls are equal in performance to boys. They are 95% sure their assumption is correct.
To analyze the data using the t-test for unpaired data, we can follow these steps:
Step 1: State the null and alternative hypotheses.
The null hypothesis (H0) is that there is no difference in test scores between the girls and boys. The alternative hypothesis (Ha) is that there is a difference in test scores between the girls and boys.
H0: μ1 = μ2
Ha: μ1 ≠ μ2
Step 2: Define the significance level.
The significance level (α) is given as 0.05 or 5%.
Step 3: Calculate the test statistic.
The t-test formula for unpaired data is given by:
t = (xbar1 - xbar2) / sqrt((s1^2/n1) + (s2^2/n2))
Where:
xbar1 and xbar2 are the sample means of the girls and boys, respectively.
s1 and s2 are the sample standard deviations of the girls and boys, respectively.
n1 and n2 are the sample sizes of the girls and boys, respectively.
Plugging in the given values:
xbar1 = 25
xbar2 = 26
s1 = 2.2
s2 = 3.4
n1 = 22
n2 = 24
Calculating the test statistic:
t = (25 - 26) / sqrt((2.2^2/22) + (3.4^2/24))
Step 4: Find the critical value.
Since the significance level is 0.05 and we have a two-tailed test, the critical value is determined by dividing the significance level by 2 and looking it up in the t-distribution table with (n1 + n2 - 2) degrees of freedom.
Since n1 = 22 and n2 = 24, the degrees of freedom will be (22 + 24) - 2 = 44.
Looking up the critical value for a 2-tailed test with 44 degrees of freedom in the t-distribution table, it is approximately 2.02.
Step 5: Make a decision.
If the calculated test statistic falls outside the critical region (t < -2.02 or t > 2.02), we reject the null hypothesis and conclude that there is a significant difference in test scores between girls and boys. If the calculated test statistic falls within the critical region, we fail to reject the null hypothesis and conclude that there is no significant difference.
Step 6: Calculate the p-value.
The p-value is the probability of observing a test statistic as extreme or more extreme than the one calculated, assuming the null hypothesis is true.
We can calculate the p-value using the t-distribution with (n1 + n2 - 2) degrees of freedom and the calculated test statistic. If the p-value is less than the significance level (α), we reject the null hypothesis.
Step 7: Make a conclusion.
Based on the calculated p-value, we can decide whether to reject or fail to reject the null hypothesis. If the p-value is less than 0.05, we reject the null hypothesis and conclude that there is a significant difference in test scores between girls and boys. If the p-value is greater than or equal to 0.05, we fail to reject the null hypothesis and conclude that there is no significant difference.
Please note that I have provided the step-by-step process for conducting the t-test for unpaired data. However, the calculations and interpretation of the results should be done using statistical software or a calculator capable of performing the t-test.
To analyze the data using the T-test for unpaired data, we need to follow a step-by-step process. Here's how you can do it:
Step 1: State the null and alternative hypotheses:
- Null Hypothesis (H0): The mean test scores for girls and boys are equal.
- Alternative Hypothesis (Ha): The mean test scores for girls and boys are not equal.
Step 2: Determine the significance level (alpha):
The given information states that the school district is 95% sure their assumption is correct. Therefore, the significance level (alpha) is 0.05.
Step 3: Calculate the test statistic:
To calculate the test statistic, we use the following formula:
t = (xbar1 - xbar2) / sqrt((s1^2 / n1) + (s2^2 / n2))
Where:
- xbar1 and xbar2 are the sample means of the test scores for girls and boys respectively.
- s1 and s2 are the standard deviations of the test scores for girls and boys respectively.
- n1 and n2 are the sample sizes for girls and boys respectively.
Plugging in the given values:
xbar1 = 25, xbar2 = 26, s1 = 2.2, s2 = 3.4, n1 = 22, n2 = 24
t = (25 - 26) / sqrt((2.2^2 / 22) + (3.4^2 / 24))
Step 4: Calculate the degrees of freedom:
The degrees of freedom can be calculated using the following formula:
df = (s1^2 / n1 + s2^2 / n2)^2 / ((s1^2 / n1)^2 / (n1 - 1) + (s2^2 / n2)^2 / (n2 - 1))
Plugging in the given values:
s1 = 2.2, n1 = 22, s2 = 3.4, n2 = 24
df = (2.2^2 / 22 + 3.4^2 / 24)^2 / ((2.2^2 / 22)^2 / (22 - 1) + (3.4^2 / 24)^2 / (24 - 1))
Step 5: Find the critical value:
Since the question states that the district is 95% sure, corresponding to a 0.05 significance level, we need to find the critical value for a two-tailed test at alpha = 0.05.
Using a T-table or a statistical software, find the critical value for the degrees of freedom calculated in Step 4.
Step 6: Make a decision:
Compare the calculated test statistic (Step 3) with the critical value (Step 5).
- If the calculated test statistic is greater than the critical value, reject the null hypothesis.
- If the calculated test statistic is less than or equal to the critical value, fail to reject the null hypothesis.
This decision will determine whether the school district's assumption (null hypothesis) is correct or not.