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Lilit [14]
3 years ago
9

Fifty draws are made at random with replacement from the box [ 0 0 1 1 1]. There are 33 ticket 1’s among the draws. The expected

value for the sum is ____ , the observed value is ____ , the chance error is ____ , and the standard error is ____ . Fill in the blanks. Hint: Chance error = ObservedOutcome – ExpectedOutcome.
Mathematics
1 answer:
Aleks04 [339]3 years ago
5 0

Answer:

- The Expected value for the sum is 30.

- The Observed sum of 50 draws is 33.

- The Chance Error on the 50 draws is 3.

- The Standard Error on 50 draws is 2.191.

Step-by-step explanation:

The box contains [0, 0, 1, 1, 1]

Using probability to predict the expected outcome.

On one draw, the probability of drawing a 0 is (2/5).

And the probability of drawing a 1 is (3/5).

Probability mass function would look like

X | P(X)

0 | 0.40

1 | 0.60

So, expected value on one draw would be

E(X) = Σ xᵢpᵢ

xᵢ = each variable

pᵢ = probability of each variable

E(X) = (0×0.40) + (1×0.60) = 0.60.

Standard error on one draw = √[Σ(xᵢ - μ)²/N]

μ = E(X) = 0.60

Σ(xᵢ - μ)² = (0 - 0.60)² + (0 - 0.60)² + (1 - 0.6)² + (1 - 0.6)² + (1 - 0.6)² = 1.20

SE = √(1.2/5) = 0.490

So, for 50 draws (with replacement),

E(50X) = 50E(X) = 50 × 0.60 = 30.

For 50 draws, standard error = √50 × 0.490 = 2.191

The expected value for the sum = 30

The observed valued for the sum = (33×1) + (17×0) = 33

Chance Error = (Observed Outcome) - (Expected Outcome) = 33 - 30 = 3

Standard error gives an idea of how large the chance error would be.

Standard error on 50 draws = 2.191

Hope this Helps!!!

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#10 i The table shows the admission costs (in dollars) and the average number of daily visitors at an amusement park each the pa
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The line of best fit is a straight line that can be used to predict the

average daily attendance for a given admission cost.

Correct responses:

  • The equation of best fit is; \underline{ \hat Y = 1,042 - 4.9 \cdot X_i}
  • The correlation coefficient is; r ≈<u> -0.969</u>

<h3>Methods by which the line of best fit is found</h3>

The given data is presented in the following tabular format;

\begin{tabular}{|c|c|c|c|c|c|c|c|c|}Cost, (dollars), x&20&21&22&24&25&27&28&30\\Daily attendance, y&940&935&940&925&920&905&910&890\end{array}\right]

The equation of the line of best fit is given by the regression line

equation as follows;

  • \hat Y = \mathbf{b_0 + b_1 \cdot X_i}

Where;

\hat Y = Predicted value of the<em> i</em>th observation

b₀ = Estimated regression equation intercept

b₁ = The estimate of the slope regression equation

X_i = The <em>i</em>th observed value

b_1 = \mathbf{\dfrac{\sum (X - \overline X) \cdot (Y - \overline Y) }{\sum \left(X - \overline X \right)^2}}

\overline X = 24.625

\overline Y = 960.625

\mathbf{\sum(X - \overline X) \cdot (Y - \overline Y)} = -433.125

\mathbf{\sum(X - \overline X)^2} = 87.875

Therefore;

b_1 = \mathbf{\dfrac{-433.125}{87.875}} \approx -4.9289

Therefore;

  • The slope given to the nearest tenth is b₁ ≈ -4.9

b_0 = \mathbf{\dfrac{\left(\sum Y \right) \cdot \left(\sum X^2 \right) - \left(\sum X \right) \cdot \left(\sum X \cdot Y\right)} {n \cdot \left(\sum X^2\right) - \left(\sum X \right)^2}}

By using MS Excel, we have;

n = 8

∑X = 197

∑Y = 7365

∑X² = 4939

∑Y² = 6782675

∑X·Y = 180930

(∑X)² = 38809

Therefore;

b_0 = \dfrac{7365 \times 4939-197 \times 180930}{8 \times 4939 - 38809} \approx \mathbf{1041.9986}

  • The y-intercept given to the nearest tenth is b₀ ≈ 1,042

The equation of the line of best fit is therefore;

  • \underline{\hat Y = 1042 - 4.9 \cdot X_i}

The correlation coefficient is given by the formula;

\displaystyle r = \mathbf{\dfrac{\sum \left(X_i - \overline X) \cdot \left(Y - \overline Y \right)}{ \sqrt{\sum \left(X_i - \overline X \right)^2 \cdot \sum \left(Y_i - \overline Y \right)^2} }}

Where;

\sqrt{\sum \left(X - \overline X \right)^2 \times \sum \left(Y - \overline Y \right)^2}  = \mathbf{446.8121}

\sum \left(X_i - \overline X \right) \times \left(Y - \overline Y\right) = \mathbf{-433.125}

Which gives;

r = \dfrac{-433.125}{446.8121}  \approx \mathbf{-0.969367213}

The correlation coefficient given to the nearest thousandth is therefore;

  • <u>Correlation coefficient, r ≈ -0.969</u>

Learn more about regression analysis here:

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