Does computer-assisted instruction help community college health students with reading difficulties to learn reading skills at a faster than normal rate? A researcher arranged for one of these students to have access to a set of computer-learning programs instead of the normal reading curriculum for one term. At the end of the term, the researcher tested her on a standardized reading ability test on which the mean for students with reading difficulties is 36 with a standard deviation of 6. The test participant scored 47. Was this result significant at the .05 level?

Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.

To determine if the result is significant at the .05 level, we need to perform a hypothesis test using the z-test statistic and compare it to the critical value.

Here's how you can approach this problem:

Step 1: State the hypotheses:
The null hypothesis (H0) is that computer-assisted instruction does not help community college health students with reading difficulties to learn reading skills at a faster rate. The alternative hypothesis (H1) is that computer-assisted instruction does help students to learn reading skills faster.

Step 2: Set the significance level:
The significance level (α) is given as .05 or 5%, which means we need to compare the test statistic to the critical value at this level.

Step 3: Calculate the test statistic:
To calculate the test statistic (z), we need the sample mean (x̄), the population mean (μ), and the standard deviation (σ).

Given:
Sample mean (x̄) = 47
Population mean (μ) = 36
Standard deviation (σ) = 6

The test statistic formula for a single sample z-test is:

z = (x̄ - μ) / (σ / √n)

Substituting the values:
z = (47 - 36) / (6 / √1)

Simplifying:
z = 11 / 6

Step 4: Compare the test statistic to the critical value:
To determine if the result is significant, we compare the calculated test statistic to the critical value at the given significance level (α = .05).

For a two-tailed test at the .05 level, the critical z-value is ±1.96.

If the calculated test statistic falls outside the critical value range, we can reject the null hypothesis.

In this case, the calculated test statistic of z = 11/6 is approximately 1.833.

Step 5: Make a conclusion:
Since the calculated test statistic (1.833) does not fall outside the critical value range (-1.96 to 1.96), we fail to reject the null hypothesis.

Therefore, we do not have enough evidence to conclude that computer-assisted instruction helps community college health students with reading difficulties to learn reading skills at a faster rate at the .05 significance level.

In summary, the result is not significant at the .05 level.