Question

Is there statistically significant evidence that the districts with smaller classes have higher average test​ scores? The t​-statistic for testing the null hypothesis is nothing. ​(Round your response to two decimal places.​) The p​-value for the test is nothing. ​(Round your response to six decimal places.​)​ Hint: Use the Excel function Norm.S.Dist to help answer this question. Is there statistically significant evidence that the districts with smaller classes have higher average test​ scores? The ▼ small p-value large p-value suggests that the null hypothesis ▼ cannot be rejected can be rejected with a high degree of confidence.​ Hence, ▼ there is there is no statistically significant evidence that the districts with smaller classes have higher average test scores.

Answer 1

Complete Question

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Answer:

The  95% confidence interval is $[670.03 , 673.97 ]$

The  test statistics is $t = 7.7$

The  p-value  is    $p-value = 0$

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  $\= y = 672.0$

   The standard deviation is  $s = 20.3$

Given that the confidence level  is 95% then the level of significance is  

   $\alpha = (100 -95 )\% = 0.05$

From the normal distribution table  the critical value of $\frac{\alpha }{2} = \frac{0.05 }{2}$ is  

    $Z_{\frac{\alpha }{2} } = 1.96$

Generally  the margin of error is mathematically represented as  

     $E = Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{n} }$

=>   $E = 1.96 * \frac{20.3}{\sqrt{408} }$

=>     $E = 1.970$

Generally the 95% confidence interval is mathematically represented as

       $\= y -E < \mu < \= y + E$

=>     $672.0 -1.970 < \mu < 672.0 +1.970$

=>     $670.03 < \mu < 673.97$

=>     $[670.03 , 673.97 ]$

From the question we are told that

   Class size                                  small                                      large

  $Avg.score(\= y)$         $\= y_1 = 683.7$   $\= y_2 = 676.0$

   $S_y$                          $S_{y_1} =20.2$    $S_{y_2} = 18.6$

   sample size                             $n_1 = 229$        $n_2 = 184$

The  null hypothesis is  $H_o : \mu_1 - \mu_2 = 0$

The alternative hypothesis is  $H_a : \mu_1 - \mu_2 > 0$

Generally the standard error for the difference in mean is mathematically represented as

       $SE = \sqrt{\frac{S_{y_1}^2 }{n_1} +\frac{S_{y_2}^2 }{n_2} }$

=>     $SE = \sqrt{20.2^2 }{229} +\frac{18.6^2 }{184_2} }$

=>     $SE = 1.913$

Generally the test statistics is mathematically represented as

      $t = \frac{\= y _1 - \= y_2 }{SE}$

=>    $t = \frac{683.7 - 676.0 }{1.913}$

=>   $t = 7.7$

Generally the p-value is mathematically represented as

    $p-value = P(t > 7.7 )$

From the  z-table

        $P(t > 7.7 ) = 0$

So

   $p-value = 0$

From the values we obtained and calculated we can see that $p-value < \alpha$

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