Respuesta :
Answer:
(1) Correct option is B.
(2) Correct option is C.
Step-by-step explanation:
The information provided is:
[tex]n_{1}=200,\ \bar x_{1}=22.7,\ s_{1} = 4.5\\n_{2}=200,\ \bar x_{2}=19.7,\ s_{2} = 4.3[/tex]
The (1 - α)% confidence interval for the difference between two mean is:
[tex]CI=\bar x_{1}-\bar x_{2}\pm t_{\alpha/2, (n_{1}+n_[2}-2}\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}} }[/tex]
The critical value of t is:
[tex]\alpha /2=0.05/2=0.025[/tex]
degrees of freedom [tex]=n_{1}+n_{2}-2=200+200-2=398[/tex]
[tex]t_{\alpha/2, n_{1}+n_{2}-2}=t_{0.025, 398}=1.96[/tex]
Compute the 95% confidence interval for the difference between two mean as follows:
[tex]CI=22.7-19.7\pm 1.96\sqrt{\frac{4.5^{2}}{200}+\frac{4.3^{2}}{200} }\\=3\pm0.8624\\=(2.1376, 3.8624)\\\approx(2.14, 3.86)[/tex]
Thus, the 95% confidence interval, (2.14, 3.86) implies that the true mean difference value is contained in this interval with probability 0.95.
Correct option is B.
The null value of the difference between means is 0.
As the value 0 is not in the interval this implies that there is a difference between the two means, concluding that priming does have an effect on scores.
Correct option is C.
