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Norma-Jean [14]

3 years ago

7

5 qt = ? gal????????

2 answers:

strojnjashka [21]3 years ago

8 0

Answer:

1.25 gal

Step-by-step explanation:

5/4 = 1.25

larisa [96]3 years ago

4 0

20 quarts. There are 4 quarts in 1 gallon. 5x4=20

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3 years ago

In a regression analysis involving 30 observations, the following estimated regressionequation was obtained.y^ =17.6+3.8x 1 −2.3

USPshnik [31]

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(a) There is a significant relationship between y and $x_1, x_2, x_3, x_4$

(b) $SSE_{(x_1 ,x_2 ,x_3 ,x_4) }= 45$

(c) $SSE_{(x_2,x_3)} = 100$

(d) $x_1$ and $x_4$ are significant

Step-by-step explanation:

Given

$y = 17.6+3.8x_1 - 2.3x_2 +7.6x_3 +2.7x_4$ --- estimated regression equation

$n = 30$

$p = 4$ --- independent variables i.e. x1 to x4

$SSR = 1760$

$SST = 1805$

$\alpha = 0.05$

Solving (a): Test of significance

We have:

$H_o :$ There is no significant relationship between y and $x_1, x_2, x_3, x_4$

$H_a :$ There is a significant relationship between y and $x_1, x_2, x_3, x_4$

First, we calculate the t-score using:

$t = \frac{SSR}{p} \div \frac{SST - SSR}{n - p - 1}$

$t = \frac{1760}{4} \div \frac{1805- 1760}{30 - 4 - 1}$

$t = 440 \div \frac{45}{25}$

$t = 440 \div 1.8$

$t = 244.44$

Next, we calculate the p value from the t score

Where:

$df = n - p - 1$

$df = 30 -4 - 1=25$

The p value when $t = 244.44$ and $df = 25$ is:

$p =0$

So:

$p < \alpha$ i.e. $0 < 0.05$

Solving (b): $SSE(x_1 ,x_2 ,x_3 ,x_4)$

To calculate SSE, we use:

$SSE = SST - SSR$

Given that:

$SSR = 1760$ ----------- $(x_1 ,x_2 ,x_3 ,x_4)$

$SST = 1805$

So:

$SSE_{(x_1 ,x_2 ,x_3 ,x_4)} = 1805 - 1760$

$SSE_{(x_1 ,x_2 ,x_3 ,x_4) }= 45$

Solving (c): $SSE(x_2 ,x_3)$

To calculate SSE, we use:

$SSE = SST - SSR$

Given that:

$SSR = 1705$ ----------- $(x_2 ,x_3)$

$SST = 1805$

So:

$SSE_{(x_2,x_3)} = 1805 - 1705$

$SSE_{(x_2,x_3)} = 100$

Solving (d): F test of significance

The null and alternate hypothesis are:

We have:

$H_o :$ $x_1$ and $x_4$ are not significant

$H_a :$ $x_1$ and $x_4$ are significant

For this model:

$y =11.1 -3.6x_2+8.1x_3$

$SSE_{(x_2,x_3)} = 100$

$SST = 1805$

$SSR_{(x_2 ,x_3)} = 1705$

$SSE_{(x_1 ,x_2 ,x_3 ,x_4) }= 45$

$p_{(x_2,x_3)} = 2$

$\alpha = 0.05$

Calculate the t-score

$t = \frac{SSE_{(x_2,x_3)}-SSE_{(x_1,x_2,x_3,x_4)}}{p_{(x_2,x_3)}} \div \frac{SSE_{(x_1,x_2,x_3,x_4)}}{n - p - 1}$

$t = \frac{100-45}{2} \div \frac{45}{30 - 4 - 1}$

$t = \frac{55}{2} \div \frac{45}{25}$

$t = 27.5 \div 1.8$

$t = 15.28$

Next, we calculate the p value from the t score

Where:

$df = n - p - 1$

$df = 30 -4 - 1=25$

The p value when $t = 15.28$ and $df = 25$ is:

$p =0$

So:

$p < \alpha$ i.e. $0 < 0.05$

<em>Hence, we reject the null hypothesis</em>

7 0

3 years ago