Two sets of 100 students each were taught to read by two different methods. After

Question

the instructions were over, a reading test given to them revealed that x = 73.4

Answer 1

To determine which method was more effective in teaching reading, we can conduct a hypothesis test.

Null Hypothesis (H0): There is no difference in reading ability between the two methods.
Alternative Hypothesis (H1): There is a difference in reading ability between the two methods.

We can use a two-sample t-test to compare the means of the two groups. Since we have a large sample size (100 students in each group), we can assume that the sample means are normally distributed.

Let's denote the mean reading test score for method 1 as μ1 and for method 2 as μ2. Given that x = 73.4, let's assume that the sample mean reading test score for method 1 is x̄1 and for method 2 is x̄2.

Since x̄1 = x̄2 = 73.4, the difference in sample means (Δ) would be 0:

Δ = x̄1 - x̄2 = 0

Now, we can calculate the standard error of the difference in sample means:

σΔ = sqrt((σ1^2/n1) + (σ2^2/n2))

Since we do not have the population standard deviation, we can estimate it using the sample standard deviations (s1 and s2) and the sample sizes (n1 = n2 = 100):

σΔ = sqrt((s1^2/n1) + (s2^2/n2))

Once we have calculated σΔ, we can then compute the t-statistic:

t = (x̄1 - x̄2) / σΔ

Next, we can determine the degrees of freedom for the t-distribution:

df = n1 + n2 - 2

Finally, we can find the p-value associated with the t-value and compare it to the significance level (α) to make a decision about the null hypothesis.

If the p-value is less than α, we reject the null hypothesis and conclude that there is a significant difference in reading ability between the two methods. If the p-value is greater than α, we fail to reject the null hypothesis and conclude that there is no significant difference in reading ability between the two methods.