Answer in Statistics and Probability for boo #249329

“bar{x_1} = 72 \nns_1 = 9.3 \nnn_1 = 13 \nnbar{x_2} = 80.2 \nns_2 = 10.1 \nnn_2 = 13”

The objective of the study is to test does the new booklet appear to be better than the standard version.

Under the claim the null and alternative hypotheses are

“H_0: mu_1 = mu_2 \nnH_1: mu_1 < mu_2 \”

α=0.05

Test-statistic

“t = frac{(bar{x_1} – bar{x_2}) -(mu_1 -mu_2)}{sqrt{s^2_{p} times (frac{1}{n_1} + frac{1}{n_2})}} \nns^2_{p} = frac{(n_1-1) times s^2_1 + (n_2-1) times s^2_2}{n_1+n_2-2} \nn= frac{(13-1) times 9.3^2 + (13-1) times 10.1^2}{13+13-2} \nn= 94.25”

Substitute the values in test statistic as shown below:

“t = frac{(72.0-80.2) -0}{sqrt{94.25(frac{1}{13} + frac{1}{13})}} \nn= -2.15”

Degrees of freedom:

“df= n_1+n_2 -2 = 13+13 -2 = 24”

At the given level of significance 0.05 and the test is left-tailed, then from t-table corresponding to 24 degrees of freedom the critical value is equal to -1.711.

Here, the calculated t value -2.15 is less than the tabulated value -1.711, reject the null hypothesis.

Hence, it can be concluded that new booklet is better than the standard version.