To determine whether there is evidence to suggest that the online system improves test scores, we can perform a two-sample t-test. The null hypothesis (H0) is that the true mean score under the pencil and paper system (μ1) is equal to the true mean score under the online system (μ2). The alternative hypothesis (H1) is that the true mean score under the online system (μ2) is greater than the true mean score under the pencil and paper system (μ1).

Here are the steps:

1. Calculate the pooled standard deviation (Sp):

Sp = sqrt[((n1-1)*s1^2 + (n2-1)*s2^2) / (n1+n2-2)]
= sqrt[((14-1)*10.2^2 + (16-1)*11.1^2) / (14+16-2)]
= sqrt[(169.2 + 180.5) / 28]
= sqrt[349.7 / 28]
= sqrt[12.49]
= 3.53

2. Calculate the standard error (SE):

SE = Sp * sqrt[(1/n1) + (1/n2)]
= 3.53 * sqrt[(1/14) + (1/16)]
= 3.53 * sqrt[0.0714 + 0.0625]
= 3.53 * sqrt[0.1339]
= 3.53 * 0.366
= 1.29

3. Calculate the t-score:

t = (x1 - x2) / SE
= (73.4 - 77.4) / 1.29
= -4 / 1.29
= -3.1

4. Compare the calculated t-score with the critical t-value from the t-distribution table. The degrees of freedom (df) is n1 + n2 - 2 = 14 + 16 - 2 = 28. For a one-tailed test at α = 0.05, the critical t-value is 1.701.

Since the calculated t-score (-3.1) is less than the critical t-value (1.701), we reject the null hypothesis. There is evidence to suggest that the online system improves test scores.

Question

To determine whether there is evidence to suggest that the online system improves test scores, we can perform a two-sample t-test. The null hypothesis (H0) is that the true mean score under the pencil and paper system (μ1) is equal to the true mean score under the online system (μ2). The alternative hypothesis (H1) is that the true mean score under the online system (μ2) is greater than the true mean score under the pencil and paper system (μ1).

Here are the steps:

1. Calculate the pooled standard deviation (Sp):

Sp = sqrt[((n1-1)*s1^2 + (n2-1)*s2^2) / (n1+n2-2)]
   = sqrt[((14-1)*10.2^2 + (16-1)*11.1^2) / (14+16-2)]
   = sqrt[(169.2 + 180.5) / 28]
   = sqrt[349.7 / 28]
   = sqrt[12.49]
   = 3.53

2. Calculate the standard error (SE):

SE = Sp * sqrt[(1/n1) + (1/n2)]
   = 3.53 * sqrt[(1/14) + (1/16)]
   = 3.53 * sqrt[0.0714 + 0.0625]
   = 3.53 * sqrt[0.1339]
   = 3.53 * 0.366
   = 1.29

3. Calculate the t-score:

t = (x1 - x2) / SE
  = (73.4 - 77.4) / 1.29
  = -4 / 1.29
  = -3.1

4. Compare the calculated t-score with the critical t-value from the t-distribution table. The degrees of freedom (df) is n1 + n2 - 2 = 14 + 16 - 2 = 28. For a one-tailed test at α = 0.05, the critical t-value is 1.701.

Since the calculated t-score (-3.1) is less than the critical t-value (1.701), we reject the null hypothesis. There is evidence to suggest that the online system improves test scores.

Knowee AI · Accepted Answer