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3 years ago

5

Four int variables, x1, x2, y1, and y2, have been declared and been given values. write an expression whose value is the differe

nce between the larger of x1 and x2 and the smaller of y1 and y2.

1 answer:

Katen [24]3 years ago

7 0

Even though we are using variables, we still know that "difference" means subtraction. So, if the larger is x1 and x2, those variables go first. The smaller, y1 and y2 will go second. So our problem will look like this:

(x1 < x2 ? x2 : x1) - (y1 < y2 ? y1 : y2)

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3 years ago

Is there statistically significant evidence that the districts with smaller classes have higher average test scores? The t-sta

Musya8 [376]

Complete Question

The complete question is shown on the first uploaded image

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 <u>p-value</u> suggests that the null hypothesis is<u> rejected </u>with a high degree of confidence. Hence <u>there is</u> 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

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Step-by-step explanation:

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2 years ago

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iogann1982 [59]

Hello,

f(n+1)=f(n)-2
f(1)=10
f(2)=8
f(3)=6
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3 years ago

Read 2 more answers