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

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
`(\alpha )/(2) = (0.05 )/(2)` is

`Z_{(\alpha )/(2) } =  1.96`

Generally the margin of error is mathematically represented as

`E  =  Z_{(\alpha )/(2) } *  (s)/(√(n) )`

=>
`E  =  1.96 *  (20.3)/(√(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{(S_(y_1)^2 )/(n_1) +(S_(y_2)^2 )/(n_2)   }`

=>
`SE =  √(20.2^2 ){229} +(18.6^2 )/(184_2)   }`

=>
`SE =  1.913`

Generally the test statistics is mathematically represented as

`t = (\= y _1 - \= y_2 )/(SE)`

=>
`t = (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