Mathematics: A consumer group is testing camp stoves. To test the heating capacity of a stove, they measure the time required to bring 2 quarts of water from 50 degrees to boiling. Two competing models are under consideration. Thirty-six stoves of each model were tested and the following results were obtained. Model 1: mean time is 11.4 and standard deviation is2.5 Model 2: mean time is 9.9 and standard deviation is 3.0 Is there any difference between the performances of these two models? {use a .05 level of

A consumer group is testing camp stoves. To test

Ответ:

grimwolf1023

14.02.2022

Mathematics

The decision rule is

Reject the null hypothesis

The conclusion is

There is sufficient evidence to show that there is a difference between the performances of these two models

The 95% confidence interval is $0.224 < \mu_1 - \mu_2 < 2.776$

Step-by-step explanation:

From the question we are told that

The sample size is n = 36

The first sample mean is $\= x_1 = 11.4$

The first standard deviation is s_1 = 2.5

The second sample mean is $\= x_2 = 9.9$

The second standard deviation is s_2 = 3.0

The level of significance is $\alpha = 0.05$

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

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

Generally the test statistics is mathematically represented as

$z = \frac{ (\= x_1 - \= x_2 ) - (\mu_1 - \mu_2 ) }{ \sqrt{ \frac{s_1^2 }{n} + \frac{s_2^2 }{ n} } }$

=> $z = \frac{ ( 11.4 - 9.9) - 0 }{ \sqrt{ \frac{2.5^2 }{36} + \frac{ 3^2 }{36 } } }$

=> z = 2.3

From the z table the area under the normal curve to the left corresponding to 2.3 is

P( Z 2.3 ) = 0.010724

Generally the p-value is mathematically represented as

p-value = 2 * P( Z 2.3 )

=> p-value = 2 * 0.010724

=> p-value = 0.02

From the value obtained we see that $p-value < \alpha$ hence

The decision rule is

Reject the null hypothesis

The conclusion is

There is sufficient evidence to show that there is a difference between the performances of these two models

Considering question b

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{s_1^2 }{n } + \frac{s_2^2}{n}}$

=> $E = 1.96 * \sqrt{ \frac{2.5^2 }{ 36 } + \frac{ 3^2}{36}}$

=> E = 1.276

Generally 95% confidence interval is mathematically represented as

$( \= x_1 - \= x_2) -E < \mu_1 - \mu_2 < ( \= x_1 - \= x_2) + E$

=> $( 11.4 - 9.9 ) -1.276 < \mu_1 - \mu_2 < ( 11.4 - 9.9 ) + 1.276$

=> $0.224 < \mu_1 - \mu_2 < 2.776$

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Ответ:

genyjoannerubiera

17.05.2023

Mathematics

1.
$y=3(x-6)^2-9 \\ \\
\hbox{x-intercept - y=0} \\
0=3(x-6)^2-9 \\
9=3(x-6)^2 \ \ \ |\div 3 \\
3=(x-6)^2 \\
\sqrt{3}=\sqrt{(x-6)^2} \\
\pm \sqrt{3}=x-6 \\
\pm \sqrt{3}+6=x \\
x=6+\sqrt{3} \ \lor \ x=6-\sqrt{3}} \\ \\
\hbox{y-intercept - x=0} \\
y=3(0-6)^2-9 \\
y=3 \times (-6)^2-9 \\
y=3 \times 36-9 \\
y=108-9 \\
y=99 \\ \\
\boxed{\hbox{x-intercepts: } 6+\sqrt{3} \hbox{ and } 6-\sqrt{3}} \\
\boxed{\hbox{y-intercept: } 99}$

2.
$y=-2(x+5)^2+3 \\ \\
\hbox{x-intercept - y=0} \\
0=-2(x+5)^2+3 \\
-3=-2(x+5)^2 \ \ \ |\div (-2) \\
\frac{3}{2}=(x+5)^2 \\
\sqrt{\frac{3}{2}}=\sqrt{(x+5)^2} \\
\pm \sqrt{\frac{3}{2}}=x+5 \\
\pm \sqrt{\frac{3}{2}}-5=x \\
x=\sqrt{\frac{3}{2}}-5 \ \lor \ x=-\sqrt{\frac{3}{2}}-5 \\ \\
\hbox{y-intercept - x=0} \\
y=-2(0+5)^2+3 \\
y=-2 \times 5^2 +3 \\
y=-2 \times 25+3 \\
y=-50+3 \\
y=-47 \\ \\ \boxed{\hbox{x-intercepts: } \sqrt{\frac{3}{2}}-5 \hbox{ and } -\sqrt{\frac{3}{2}}-5} \\ \boxed{\hbox{y-intercept: } -47}$

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