Answer:
There is not enough statistical evidence to state that the mean score on one-tailed hypothesis test questions is higher than the mean score on two-tailed hypothesis test questions.
Step-by-step explanation:
To solve this problem, we run a hypothesis test about the difference of population means.
[tex]$Sample mean $\bar X: \bar X=7.79$\\Sample mean $\bar Y: \bar Y=7.64$\\Population variance $S^2_X: S^2_X=1.06^2$\\Population variance $S^2_Y: S^2_Y=1.31^2$\\Sample size $n_X=80$\\Sample size $n_Y=80$\\Significance level $\alpha=0.05$\\Z criticals values (for a 0.05 significance)\\(Left tail test) Z_{1-\alpha}=Z_{0.95}=-1.64485\\($Right tail test) Z_{\alpha}=Z_{0.05}=1.64485\\($Two-tailed test) $Z_{1-\alpha/2}=Z_{0.975}=-1.95996$ and $ Z_{\alpha/2}=Z_{0.025}=1.95996\\\\[/tex]
The appropriate hypothesis system for this situation is:
[tex]H_0:\mu_X-\mu_Y=0\\H_a:\mu_X-\mu_Y > 0\\\\$Difference of means in the null hypothesis is:\\\mu_X-\mu_Y=M_0=0\\\\$The test statistic is $Z=\frac{[( \bar X-\bar Y)-M_0]}{\sqrt{\frac{\sigma^2_X}{n_X}+\frac{\sigma^2_Y}{n_Y}}}\\[/tex]
[tex]$The calculated statistic is Z_c=\frac{[(7.79-7.64)-0]}{\sqrt{\frac{1.06^2}{80}+\frac{1.31^2}{80}}}=0.79616\\\\p-value = P(Z \geq Z_c)=0.42594\\\\[/tex]
Since, the calculated statistic [tex]Z_c[/tex] is less than critical [tex]Z_{\alpha}[/tex], the null hypothesis do not should be rejected. There is not enough statistical evidence to state that the mean score on one-tailed hypothesis test questions is higher than the mean score on two-tailed hypothesis test questions.
