Answer :
Complete Question
The complete question is shown on the first uploaded image
Answer:
The 95% confidence interval is [tex] [670.03 , 673.97 ] [/tex]
The test statistics is [tex]t = 7.7 [/tex]
The p-value is [tex]p-value = 0[/tex]
The p-value suggests that the null hypothesis is rejected with a high degree of confidence. Hence there is statistically significant evidence that the districts with smaller classes have higher average test score
Step-by-step explanation:
From the question we are told that
The sample size is n = 408
The sample mean is [tex]\= y = 672.0[/tex]
The standard deviation is [tex]s = 20.3[/tex]
Given that the confidence level is 95% then the level of significance is
[tex]\alpha = (100 -95 )\% = 0.05[/tex]
From the normal distribution table the critical value of [tex]\frac{\alpha }{2} = \frac{0.05 }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{n} }[/tex]
=> [tex]E = 1.96 * \frac{20.3}{\sqrt{408} }[/tex]
=> [tex]E = 1.970[/tex]
Generally the 95% confidence interval is mathematically represented as
[tex]\= y -E < \mu < \= y + E[/tex]
=> [tex] 672.0 -1.970 < \mu < 672.0 +1.970[/tex]
=> [tex] 670.03 < \mu < 673.97[/tex]
=> [tex] [670.03 , 673.97 ] [/tex]
From the question we are told that
Class size small large
[tex]Avg.score(\= y)[/tex] [tex]\= y_1 = 683.7[/tex] [tex]\= y_2 = 676.0[/tex]
[tex]S_y[/tex] [tex]S_{y_1} =20.2[/tex] [tex]S_{y_2} = 18.6[/tex]
sample size [tex]n_1 = 229[/tex] [tex]n_2 = 184[/tex]
The null hypothesis is [tex]H_o : \mu_1 - \mu_2 = 0[/tex]
The alternative hypothesis is [tex]H_a : \mu_1 - \mu_2 > 0[/tex]
Generally the standard error for the difference in mean is mathematically represented as
[tex]SE = \sqrt{\frac{S_{y_1}^2 }{n_1} +\frac{S_{y_2}^2 }{n_2} }[/tex]
=> [tex]SE = \sqrt{20.2^2 }{229} +\frac{18.6^2 }{184_2} }[/tex]
=> [tex]SE = 1.913[/tex]
Generally the test statistics is mathematically represented as
[tex]t = \frac{\= y _1 - \= y_2 }{SE}[/tex]
=> [tex]t = \frac{683.7 - 676.0 }{1.913}[/tex]
=> [tex]t = 7.7 [/tex]
Generally the p-value is mathematically represented as
[tex]p-value = P(t > 7.7 )[/tex]
From the z-table
[tex]P(t > 7.7 ) = 0[/tex]
So
[tex]p-value = 0[/tex]
From the values we obtained and calculated we can see that [tex]p-value < \alpha[/tex]
This mean that
The p-value suggests that the null hypothesis is rejected with a high degree of confidence. Hence there is statistically significant evidence that the districts with smaller classes have higher average test score
