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Novosadov [1.4K]

2 years ago

5

A new movie theater has a deal for watching lots of movies. You pay 20 dollars at the beginning of the month and then you pay ju

st 4 dollars each time you want to see a movie. How much would it cost you if you 9 movies with this deal?

1 answer:

fenix001 [56]2 years ago

5 0

20 + 4 * 9

20 + 36

$56

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Answer:

1. The least squares regression is y = -0.1015·x + 6.51

2. The independent variable is b) age

Please see attached table

Step-by-step explanation:

The least squares regression formula is given as follows;

$\dfrac{\sum_{i = 1}^{n}(x_{i} - \bar{x})\times \left (y_{i} - \bar{y} \right ) }{\sum_{i = 1}^{n}(x_{i} - \bar{x})^{2}}$

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$\bar x$ = 24

$\bar y$ = 4

$\Sigma (x_i - \bar x) (y_i - \bar y)$ = -79

$\Sigma (x_i - \bar x)^2$ = 778

$\therefore \hat \beta =\dfrac{\sum_{i = 1}^{n}(x_{i} - \bar{x})\times \left (y_{i} - \bar{y} \right ) }{\sum_{i = 1}^{n}(x_{i} - \bar{x})^{2}} = \frac{-79}{778} = -0.1015$

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∴ α = y -0.1015·x = 6 - (-0.1015 × 5) = 6.51

The least squares regression is thus;

y = -0.1015·x + 6.51

2. The independent variable is the age b)

3. Steps to create an ANOVA table with α = 0.05

The overall mean = (43 + 30 + 22 + 20 + 5 + 1 + 6 + 4 + 3 + 6 )/10 = 14

There are 2 different treatment = $df_{treat} = 2 - 1 = 1$

There are 10 different treatment measurement = $df_{tot} = 10 - 1 = 9$

$df_{res} = 9 - 1 = 8$

$df_{treat} + df_{res} = df_{tot}$

The estimated effects are;

$\hat A_1 = 24 - 14 = 10$

$\hat A_2 = 4 - 14 = -10$

$SS_{treat} = 10^2 \times 5 + (-10)^2 \times 5 =1000$

$\sum_{i}\SS_{row}_i = \sum_{i}\sum_{j} (y_{ij} - \bar y)= [(1 - 4)^2 + (6 - 4)^2 + (4 - 4)^2 + (3 - 4)^2 + (6 - 4)^2] = 18$

$\sum_{i} S S_{row}_i = \sum_{i}\sum_{j} (y_{ij} - \bar y) ^2= [(43 - 24)^2 + (30 - 24)^2 + (22 - 24)^2 + (20 - 24)^2 + (5 - 24)^2] = 778$

$S S_{res} = \sum_{i} S S_{row}_i = 778 + 18 = 796$

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$MS_{res} = \dfrac{SS_{res} }{df_{res} } = \dfrac{796}{8} = 99.5$

F- value is given by the relation;

$F = \dfrac{MS_{treat} }{MS_{res} } = \frac{1000}{99.5} = 10.05$

We then look for the critical values at degrees of freedom 1 and 8 at α = 0.05 on the F-distribution tables 5.3177

Hence; $F = 10.05 > F_{1,8}^{Krit}(5\%) = 5.3177$ , we reject the null hypothesis.

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