To draw a conclusion based on this result, we can perform a hypothesis test. The null hypothesis (\(H_0\)) would be that the new reading program has no effect, meaning the mean reading speed remains the same (\(\mu = 91\) wpm). The alternative hypothesis (\(H_1\)) would be that the new reading program is effective, leading to an increase in the mean reading speed (\(\mu > 91\) wpm).
Given the sample size (\(n = 18\)), sample mean (\(\bar{x} = 93.9\) wpm), population mean (\(\mu = 91\) wpm), and population standard deviation (\(\sigma = 10\) wpm), we can perform a one-sample z-test.
The test statistic (\(z\)) can be calculated as:
\[z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\]
Substituting the given values:
\[z = \frac{93.9 - 91}{\frac{10}{\sqrt{18}}} \approx 1.16\]
Next, we need to find the p-value associated with this z-score. This p-value represents the probability of obtaining a sample mean at least as extreme as 93.9 wpm, assuming the null hypothesis is true.
Consulting a standard normal distribution table or using a statistical software, we find that the p-value for \(z \approx 1.16\) is approximately 0.123.
Since this p-value is greater than common significance levels (e.g., 0.05), we fail to reject the null hypothesis. This means that we do not have enough evidence to conclude that the new reading program has a significant effect on the mean reading speed of second grade students.
In practical terms, based on this result, it seems that the new reading program did not lead to a statistically significant increase in the mean reading speed. Other factors might be influencing the results, or a longer period of observation may be needed to see a noticeable effect.