Do children from lower socioeconomic status (SES) neighborhoods have lower than average test-taking skills? A researcher administered a standard measure of test-taking skills to a randomly chosen child from a low SES neighborhood and found him to have a score of 38. The average on this measure for the population in general is 50 with a standard deviation of 10. What should you conclude about whether those in low SES neighborhoods have lower test-taking ability? (Use the .01 significance level.)
Use same process as indicated in your following post.
To determine whether children from lower socioeconomic status (SES) neighborhoods have lower than average test-taking skills based on the given information, we can perform a hypothesis test.
Step 1: State the hypotheses:
The null hypothesis (H0): Children from lower SES neighborhoods have the same test-taking skills as the average population.
The alternative hypothesis (Ha): Children from lower SES neighborhoods have lower test-taking skills than the average population.
Step 2: Set the significance level:
The significance level (α) is given as .01. This represents the probability of rejecting the null hypothesis when it is true. In other words, it determines how confident we want to be in our conclusion.
Step 3: Determine the test statistic:
In this case, we need to calculate the z-score, which tells us how many standard deviations away the child's test score is from the population average. The formula for calculating the z-score is:
z = (x - μ) / (σ / √n)
Where:
x = the child's test score (38)
μ = the population average (50)
σ = the population standard deviation (10)
n = 1 (as only one child's score is given)
Plugging in the values, we get:
z = (38 - 50) / (10 / √1)
Step 4: Calculate the z-score:
z = -12 / 10 = -1.2
Step 5: Determine the p-value:
Now we need to find the p-value associated with the calculated z-score. The p-value is the probability of obtaining a test statistic as extreme as or more extreme than the one observed, assuming the null hypothesis is true. We can look up this value in a standard normal distribution table or use statistical software.
Using a standard normal distribution table, we find that the p-value corresponding to a z-score of -1.2 is approximately 0.1151.
Step 6: Make a decision:
Compare the p-value to the significance level (α) to make a decision. If the p-value is less than α, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, the p-value (0.1151) is greater than the significance level (0.01). Therefore, we fail to reject the null hypothesis.
Step 7: Conclusion:
Based on the given information and the hypothesis test, we fail to find evidence to conclude that children from lower SES neighborhoods have lower test-taking ability.