Question

Does memory training help? strategy 1: you will be given 3 minutes to memorize as many as possible by just rereading the words. You will record as many words as you can remember. Strategy 2: You will then be given another list of words to memorize for 3 minutes using a memorization strategy. You will record as many words as you can remember. Suppose that 32 adults were randomly selected to try both memory strategies. Each adult calculated the difference in the # words (Strategy 2- Strategy 1). The mean of the differences was X bar diff=1.5 and the standard deviation of the differences is Sdiff = 2.7. Do these data give convincing evidence that strategy 2 improves memory at the a=0.05 significance level? a. State the null and alternative hypotheses. b. Are the requirements for conducting a hypothesis test satisfied? c. Compute the test statistic d. Approximate the P-value. Make a decision. e. State the conclusion

Answer 1

Step-by-step explanation:

a. The null hypothesis (H0) for this hypothesis test is that there is no improvement in memory between Strategy 1 and Strategy 2. The alternative hypothesis (Ha) is that Strategy 2 improves memory.

b. Before conducting a hypothesis test, we need to ensure that the requirements are satisfied. These requirements include:

1. Random sampling: The problem states that the 32 adults were randomly selected, so this requirement is satisfied.

2. Independence: It is assumed that the responses of one adult do not affect the responses of another adult. If the adults were selected independently, this requirement is satisfied.

3. Normally distributed population or large sample size: The problem does not mention the sample size or whether the population is normally distributed. However, with a sample size of 32, we can use the Central Limit Theorem to approximate the sampling distribution of the mean difference as approximately normal, assuming the differences are not extremely skewed.

c. To compute the test statistic, we need to calculate the t-value. The formula for the t-value is:

t = (Xbar_diff - 0) / (Sdiff / sqrt(n))

where Xbar_diff is the sample mean difference, Sdiff is the standard deviation of the differences, and n is the sample size.

In this case, Xbar_diff = 1.5, Sdiff = 2.7, and n = 32.

Plugging in these values, we get:

t = (1.5 - 0) / (2.7 / sqrt(32))

d. After calculating the t-value, we need to approximate the p-value. The p-value represents the probability of observing a sample mean difference as extreme as the one calculated (or more extreme) under the assumption that the null hypothesis is true.

To approximate the p-value, we need to use a t-distribution table or a statistical calculator. We can compare the t-value we calculated to the critical t-value for a significance level of 0.05 (since a = 0.05 was given in the question). If the t-value is beyond the critical t-value, the p-value will be less than 0.05, indicating strong evidence against the null hypothesis.

e. Based on the calculated p-value and the chosen significance level of 0.05, we make a decision. If the p-value is less than 0.05, we reject the null hypothesis and conclude that there is convincing evidence that Strategy 2 improves memory. If the p-value is greater than or equal to 0.05, we fail to reject the null hypothesis and do not have convincing evidence that Strategy 2 improves memory.

In summary:

a. H0: Strategy 2 does not improve memory (no improvement)

Ha: Strategy 2 improves memory

b. The requirements for conducting a hypothesis test are satisfied.

c. Compute the t-value using the formula:

t = (1.5 - 0) / (2.7 / sqrt(32))

d. Approximate the p-value by comparing the calculated t-value to the critical t-value for a significance level of 0.05.

e. Make a decision based on the p-value. If the p-value is less than 0.05, reject the null hypothesis and conclude that there is convincing evidence that Strategy 2 improves memory. Otherwise, fail to reject the null hypothesis.