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avanturin [10]
3 years ago
9

In many fast food restaurants, there is a strong correlation between a menu item's fat content (measured in grams) and its calor

ie content. We want to investigate this relationship. Using all of the food menue items at a well known fast food restuarant, the fat content and calorie contents were measured. We decide to fit the LSRL to the data, with fat content (x) as the explanatory variable and (y) as the response variable. One of the menu items is a hamburger with 107 grams of fat and 1410 calories.
r = correlation between x and y = 0.979
x bar = mean of the values of x = 40.35 grams
y bar = mean of values of y = 662.88 calories
sx = standard deviation of the values of x = 27.99 grams
sy = standard devation of the values of y = 324.90 calories
1) The slope of the least-squares regressin line is:_______.
a. 11.36
b. 11.36
c. 0.979
d. 16.08
2) The intercept of the least-squares regression line is:__________.
a. 204.50
b. 662.88
c. 662.88
d. none of the above
3) Refer to teh example data poin (107 grams, 1410 calories), what is the residual corresponding to this observation?
a. 10 calories
b. 10 grams
c. 10 calories
d. 10 grams
Mathematics
1 answer:
Kruka [31]3 years ago
8 0

Answer:

1. The slope of the least-squares regressin line is 11.364

2. The intercept of the least-squares regression line is 204.343

3. The residual corresponding to this observation is -10 calories

Step-by-step explanation:

1. In order to calculate the slope of the least-squares regressin line we would have to use the following formula:

slope of the least-squares regressin line=r*sy/sx

Therefore, slope=0.979*324.90/27.99

slope=11.364

2. In order to calculate the intercept of the least-squares regression line we would have to use the following formula:

intercept of the least-squares regression line=ybar - slope*xbar

intercept=662.88 - 11.364*40.35

intercept=204.343

3. In order to calculate the the residual corresponding to this observation we would have to use the following formula:

residual = observed y - predicted y

predicted y=11.364*x + 204.343

if the x value is 7, therefore predicted y=11.364*107 + 204.343

predicted y=1,420

Therefore, residual=1410 calories - 1420 calories

residual=-10 calories

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Construct a 90% confidence interval for μ1-μ2 with the sample statistics for mean calorie content of two​ bakeries' specialty pi
DIA [1.3K]

Answer:

The 90% confidence interval for the difference in mean (μ₁ - μ₂) for the two bakeries is; (<u>49</u>) < μ₁ - μ₂ < (<u>289)</u>

Step-by-step explanation:

The given data are;

Bakery A

\overline x_1<em> </em>= 1,880 cal

s₁ = 148 cal

n₁ = 10

Bakery B

\overline x_2<em> </em>= 1,711 cal

s₂ = 192 cal

n₂ = 10

\left (\bar{x}_1-\bar{x}_{2}  \right ) - t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}< \mu _{1}-\mu _{2}< \left (\bar{x}_1-\bar{x}_{2}  \right ) + t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}

df = n₁ + n₂ - 2

∴ df = 10 + 18 - 2 = 26

From the t-table, we have, for two tails, t_c = 1.706

\hat{\sigma} =\sqrt{\dfrac{\left ( n_{1}-1 \right )\cdot s_{1}^{2} +\left ( n_{2}-1 \right )\cdot s_{2}^{2}}{n_{1}+n_{2}-2}}

\hat{\sigma} =\sqrt{\dfrac{\left ( 10-1 \right )\cdot 148^{2} +\left ( 18-1 \right )\cdot 192^{2}}{10+18-2}}= 178.004321469

\hat \sigma ≈ 178

Therefore, we get;

\left (1,880-1,711  \right ) - 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}< \mu _{1}-\mu _{2}< \left (1,880-1,711  \right ) + 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}

Which gives;

169 - \dfrac{75917\cdot \sqrt{35} }{3,750} < \mu _{1}-\mu _{2}< 169 + \dfrac{75917\cdot \sqrt{35} }{3,750}

Therefore, by rounding to the nearest integer, we have;

The 90% C.I. ≈ 49 < μ₁ - μ₂ < 289

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