SOLUTION: An extract of a study conducted about the effects of a special class designed to aid students with verbal skills is given below: Each child was given a verbal skills test twice, bo
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Question 1185626: An extract of a study conducted about the effects of a special class designed to aid students with verbal skills is given below: Each child was given a verbal skills test twice, both before and after completing a 4-week period in the class. Let Y = score on exam at time 2-score on exam at time 1. Hence, if the population mean for Y is equal to 0, the class has no effect, on the average. For the four children in the study, the observed values of Y are 8-5 =3, 10-3 =7, 5-2=3, and 7-4=3 (e.g. for the first child, the scores were 5 on exam 1 and 8 on exam 2, so Y = 8-5=3). It is planned to test the null hypothesis of no effect against the alternative hypothesis that the effect is positive,based on the following results from a statistical software package:
Variable. Number of cases. Mean. Standard Deviation. Standard Error
Y. 4. 4.000. 2.000. 1.000
(a) Set up the null and alternative hypotheses
(b) Calculate the test statistics, and indicate whether the p-value was below 0.05, based on using the appropriate table
(c) Make a decision, using alpha = .05. Interpret.
(d) If the decision in (c) is actually (unknown to us) incorrect, what type of error has been made? What could you do to reduce the chance of that type of error?
(e) True or false? When we make a decision using alpha = .05, this means that if the special class is truly beneficial, there is only a 5% chance that we will conclude that it is not beneficial
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Here's the analysis of the study on the verbal skills class:
**(a) Null and Alternative Hypotheses:**
* **Null Hypothesis (H₀):** The class has no effect. μ_Y = 0
* **Alternative Hypothesis (H₁):** The class has a positive effect. μ_Y > 0 (This is a one-tailed test.)
**(b) Test Statistic and p-value:**
Since the sample size is small (n=4) and the population standard deviation is unknown, we use a t-test.
* **Test Statistic:**
t = (sample mean - hypothesized mean) / (standard error)
t = (4 - 0) / 1 = 4
* **p-value:**
We need to consult a t-distribution table with degrees of freedom (df) = n - 1 = 4 - 1 = 3. Look for the one-tailed probability associated with a t-value of 4.
A t-table will likely not have a value for t=4 when df = 3. This indicates that the p-value is extremely small, much less than 0.05. Statistical software would give a more precise p-value, but it will be very close to zero.
**(c) Decision and Interpretation:**
* **Decision:** Since the p-value is less than α = 0.05, we reject the null hypothesis.
* **Interpretation:** There is sufficient evidence to conclude that the special class has a positive effect on verbal skills.
**(d) Type I Error and Reduction:**
* **Type I Error:** If our decision in (c) is incorrect, it means we rejected the null hypothesis when it was actually true. In this context, it would mean we concluded the class had a positive effect when it actually had no effect. This is also called a "false positive".
* **Reducing Type I Error:** The most common way to reduce the chance of a Type I error is to decrease the significance level (alpha). For example, using α = 0.01 instead of 0.05 would make it harder to reject the null hypothesis. However, this increases the chance of a Type II error (failing to reject a false null hypothesis). Another way to reduce the chance of Type I error is to increase the sample size, if possible.
**(e) True or False?**
False. A significance level of α = 0.05 means that *if the null hypothesis is true* (the class has no effect), there is a 5% chance that we will incorrectly reject it. It does *not* mean that if the class is beneficial, there's only a 5% chance we'll conclude it's not. The latter statement confuses the null hypothesis with the alternative hypothesis.
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