using the following information, determine if the differences between the means are significant medical terminolgy class. Group A = number of word elements remembered by students using flash cards, group B= number of word elements remembered by students not using flash cards. group A, mean = 75; sd = 2.0 ; n = 10. Group B mean = 50, sd = 3.5, n = 10 The null hypothesis: There is no difference between the groups. H0: µ1 - µ2 = 0 (the difference between the means is zero) (t = 2.50, df= 18,p < 0.05
what are the answer choices
No choices given. I am at a complete loss at where to begin with this problem.
t with 18 df at .05 = 2.101
What does that tell you?
P <0.05,
It is statistically significant at the .05 level.
To determine if the difference between the means of Group A and Group B is statistically significant in the medical terminology class, we can perform a t-test.
Here's how to do it step by step:
Step 1: Set up the Null Hypothesis (H0) and Alternative Hypothesis (H1):
- Null Hypothesis (H0): There is no difference between the means of Group A and Group B. µ1 - µ2 = 0.
- Alternative Hypothesis (H1): There is a difference between the means of Group A and Group B. µ1 - µ2 ≠ 0 (two-tailed test).
Step 2: Determine the significance level (α):
- The significance level, often denoted as α, is the threshold at which we will reject the null hypothesis. In this case, the significance level is given as p < 0.05, which means we will reject the null hypothesis if the p-value is less than 0.05.
Step 3: Calculate the t-value:
- The t-value is a measure of how far the sample mean (difference between groups) is from the hypothesized population mean (0 in this case). It is calculated using the formula:
t = (mean1 - mean2) / sqrt((sd1^2/n1) + (sd2^2/n2))
where mean1 and mean2 are the means of Group A and Group B, sd1 and sd2 are their respective standard deviations, and n1 and n2 are the number of observations in each group.
Given the following information:
Group A mean = 75, sd = 2.0, n = 10
Group B mean = 50, sd = 3.5, n = 10
Plugging the values into the formula:
t = (75 - 50) / sqrt((2.0^2/10) + (3.5^2/10))
t = 25 / sqrt(0.4 + 1.225)
t ≈ 25 / sqrt(1.625)
t ≈ 25 / 1.276
t ≈ 19.59
Step 4: Calculate the degrees of freedom (df):
- The degrees of freedom is calculated as (n1-1) + (n2-1), where n1 and n2 are the number of observations in each group.
df = (10-1) + (10-1) = 18
Step 5: Determine the p-value:
- The p-value represents the probability of obtaining the observed t-value (or more extreme) assuming that the null hypothesis is true. To find the p-value, you can use statistical software, a t-distribution table, or an online calculator. The p-value for the given t-value of 19.59 and df of 18 is practically zero (p < 0.001).
Step 6: Compare the p-value to the significance level:
- Since the p-value (p < 0.001) is less than the significance level (p < 0.05), we reject the null hypothesis.
Conclusion:
Based on the analysis, we can conclude that there is a significant difference between the means of Group A and Group B in the medical terminology class (p < 0.05).