A matched pairs design was used to compare test scores in a statistics course. Twenty-four students were chosen randomly from a large class. For each chosen student, their test 1 and test 2 scores were recorded. A 95% confidence interval for the difference in test scores was constructed, along with the corresponding 2-sided p-value was given as $[-0.70, 10.45]$, and the 2-sided p-value $= 0.084$.
How does one determine whether or not this result is statistically significant at the 5% level of significance?
Also, if a $90%$ confidence interval was constructed instead of a 95% one, would it contain $0$?
Would appreciate some clues.
1) For testing $H_0: \delta = 0$ vs. $H_a: \delta \ne 0,$ where the population difference in test 1 and test 2 scores is $\mu_1 - \mu_2 = \delta,$ one does not reject $H_0$ at level 5% if the two-sided P-value exceeds 5%. Because your P-value is $0.084 > 0.05$ you do not reject.
2) Another criterion for failing to reject in (1) is that the 95% confidence interval (CI) contains $0.$ The CI can be considered an interval of 'acceptable' values of $\delta.$
3) The 95% CI in (2) is of the form $\bar D \pm t^*s_D/\sqrt{n};$ where $\bar D$ is the sample mean of the observed differences $D_i = X_i - Y_i,$ where $X_i$ and $Y_i$ the $i$th students difference in test 1 and test 2 scores; $s_D$ is the the sample standard deviation of the $D_i$; $n = 24;$ and $t^* = 2.069$ cuts 2.5% of the probability from the upper tail of Student's t distribution with $n - 1 = 23$ degrees of freedom.
For the same data, a 90% CI would be the same, except that you would use $t^* = 1.714,$ which cuts 5% of the probability from the upper tail of the same distribution. I will leave it to you to determine what difference the change from $t^* = 2.069$ to $t^* = 1.714$ would make in the width of the CI.
Note: You should look at printed tables of t distributions (or use software) to verify the values of $t^*$ quoted above.