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Simora [160]
3 years ago
7

Intrigued by a 2013 study at the University of Nebraska that suggested marijuana smokers may be thinner than other adults, a gro

up of students did a project to explore the relationship between body mass index (BMI) and amount of time spent under the influence of marijuana (measured in hours per week). Based on a random sample of 33 students at their university, they used BMI as the response variable. The equation of the least-squares regression line is: hours per month under influence = 49.2−1.15 BMI. Also, r2=0.134 .
Mathematics
1 answer:
Reptile [31]3 years ago
8 0

this question is incomplete, the complete question is:

1. is this model effective

2. what is the correlation coefficient for this data.

3. for a student with a bmi of 25, what is the predicted number of hours under the influence.

Step-by-step explanation:

1. first of all this model is not effective because we have r² as 0.134. this tells us that only 13.4 percent of the of the variations that exist in this data has been explained by the model

1. we get the correlation coefficient by

+-\sqrt{r^{2} }

the regression slope coefficient has a negative sign. this is what we would use in calculating the correlation coefficient.

-\sqrt{r^{2} }

= -√0.134

= -0.366

therefore the correlation coefficient is -0.366

2. to get the number of hours under the influence with a bmi of 25

the equation is

49.2-1.15bmi

= 49.2-1.15(25)

= 49.2-28.75

= 20.45

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2 years ago
Solid fats are more likely to raise blood cholesterol levels than liquid fats. Suppose a nutritionist analyzed the percentage of
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Answer:

t = 31.29

Step-by-step explanation:

Given

\begin{array}{ccccccc}{Stick} & {25.8} & {26.9} & {26.2} & {25.3} & {26.7}& {26.1} \ \\ {Liquid} & {16.9} & {17.4} & {16.8} & {16.2} & {17.3}& {16.8} \ \end{array}

Required

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Let the dataset of stick be A and Liquid be B.

We start by calculating the mean of each dataset;

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n, in both datasets in 6

For A

\bar x_A =\frac{25.8+26.9+26.2+25.3+26.7+26.1}{6}

\bar x_A =\frac{157}{6}

\bar x_A =26.17

For B

\bar x_B =\frac{16.9+17.4+16.8+16.2+17.3+16.8}{6}

\bar x_B =\frac{101.4}{6}

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Next, calculate the sample standard deviation

This is calculated using:

s = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}

For A

s_A = \sqrt{\frac{\sum(x - \bar x_A)^2}{n-1}}

s_A = \sqrt{\frac{(25.8-26.17)^2+(26.9-26.17)^2+(26.2-26.17)^2+(25.3-26.17)^2+(26.7-26.17)^2+(26.1-26.17)^2}{6-1}}

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s_A = \sqrt{0.34268}

s_A = 0.5854  

For B

s_B = \sqrt{\frac{\sum(x - \bar x_B)^2}{n-1}}

s_B = \sqrt{\frac{(16.9 - 16.9)^2+(17.4- 16.9)^2+(16.8- 16.9)^2+(16.2- 16.9)^2+(17.3- 16.9)^2+(16.8- 16.9)^2}{6-1}}

s_B = \sqrt{\frac{0.92}{5}}

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Calculate the pooled variance

S_p^2 = \frac{(n_A - 1)*s_A^2 + (n_B - 1)*s_B^2}{(n_A+n_B-2)}

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S_p^2 = \frac{2.6336708}{10}

S_p^2 = 0.2634

Lastly, calculate the test statistic using:

t = \frac{(\bar x_A - \bar x_B) - (\mu_A - \mu_B)}{\sqrt{S_p^2/n_A +S_p^2/n_B}}

We set

\mu_A = \mu_B

So, we have:

t = \frac{(\bar x_A - \bar x_B) - (\mu_A - \mu_A)}{\sqrt{S_p^2/n_A +S_p^2/n_B}}

t = \frac{(\bar x_A - \bar x_B) }{\sqrt{S_p^2/n_A +S_p^2/n_B}}

The equation becomes

t = \frac{(26.17 - 16.9) }{\sqrt{0.2634/6 +0.2634/6}}

t = \frac{9.27}{\sqrt{0.0878}}

t = \frac{9.27}{0.2963}

t = 31.29

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