CALCU-1. A researcher wants to investigate the eff

Solution by Steps

step 1

Identify the given correlation matrix and the standard deviations and means for each variable

step 2

Use the formula for standardized regression coefficients:

\hat{\beta}_i^* = \frac{\text{cor}(Y, X_i)}{\text{cor}(X_i, X_i)}

For

\hat{\beta}_1^*

:

\hat{\beta}_1^* = \frac{-0.815}{1.00} = -0.815

For

\hat{\beta}_2^*

:

\hat{\beta}_2^* = \frac{0.028}{1.00} = 0.028

step 3

Write the fitted standardized regression function:

\hat{Y}^* = \hat{\beta}_1^* X_1^* + \hat{\beta}_2^* X_2^*

Substitute the values:

\hat{Y}^* = -0.815 X_1^* + 0.028 X_2^*

step 4

Interpret

\hat{\beta}_2^*

when

X_1

is held constant:

\hat{\beta}_2^* = 0.028

This means that for a one standard deviation increase in

X_2

,

Y

increases by 0.028 standard deviations, holding

X_1

constant

step 5

Transform the fitted standardized regression function back into the original variables using the means and standard deviations:

Y = \bar{Y} + s_Y \hat{Y}^*

X_1 = \bar{X}_1 + s_1 X_1^*

X_2 = \bar{X}_2 + s_2 X_2^*

Substitute the values:

Y = 0.613 + 0.1243(-0.815 \frac{X_1 - 0.4}{0.1777} + 0.028 \frac{X_2 - 4.25}{1.518})

Answer

The standardized regression coefficients are

\hat{\beta}_1^* = -0.815

and

\hat{\beta}_2^* = 0.028

. The fitted standardized regression function is

\hat{Y}^* = -0.815 X_1^* + 0.028 X_2^*

. When

X_1

is held constant,

\hat{\beta}_2^* = 0.028

indicates that for a one standard deviation increase in

X_2

,

Y

increases by 0.028 standard deviations. The transformed regression function in original variables is:

Y = 0.613 + 0.1243(-0.815 \frac{X_1 - 0.4}{0.1777} + 0.028 \frac{X_2 - 4.25}{1.518})

Key Concept

Standardized Regression Coefficients

Explanation

Standardized regression coefficients measure the strength and direction of the relationship between each predictor variable and the response variable, standardized by the standard deviations of the variables.

Solution by Steps

step 1

To calculate the adjusted

R

-square, we use the formula:

R_{\text{adj}}^2 = 1 - \left( \frac{(1 - R^2)(n - 1)}{n - k - 1} \right)

step 2

From Table 1, obtain

R^2

,

n

(sample size), and

k

(number of predictors)

step 3

Substitute the values into the formula and solve for

R_{\text{adj}}^2

Part (b)

step 1

To obtain the value of

\operatorname{SSR}(X_1, X_2)

, use the formula:

\operatorname{SSR} = \sum (\hat{Y} - \bar{Y})^2

step 2

From Table 1, find the necessary values for

\hat{Y}

and

\bar{Y}

step 3

Calculate

\operatorname{SSR}(X_1, X_2)

and interpret the result

Part (c)

step 1

To test for a linear relationship, use the F-test:

F = \frac{(R^2 / k)}{((1 - R^2) / (n - k - 1))}

step 2

From Table 1, obtain

R^2

,

n

, and

k

step 3

Calculate the F-statistic and compare it to the critical value from the F-distribution table

step 4

Conclude whether there is a significant linear relationship

Part (d)

step 1

To test whether

X_3

can be dropped, use the partial F-test:

F = \frac{(\operatorname{SSR}_{\text{full}} - \operatorname{SSR}_{\text{reduced}}) / (p - q)}{\operatorname{SSR}_{\text{full}} / (n - p - 1)}

step 2

From Table 2, obtain

\operatorname{SSR}_{\text{full}}

and

\operatorname{SSR}_{\text{reduced}}

step 3

Calculate the F-statistic and compare it to the critical value

step 4

Conclude whether

X_3

can be dropped

Part (e)

step 1

To obtain the partial correlation

R_{Y3 \mid 12}^2

, use the formula:

R_{Y3 \mid 12}^2 = \frac{R_{Y3}^2 - R_{Y1}^2 - R_{Y2}^2 + 2R_{Y1}R_{Y2}R_{12}}{1 - R_{12}^2}

step 2

From Table 2, obtain the necessary correlation coefficients

step 3

Substitute the values into the formula and solve for

R_{Y3 \mid 12}^2

step 4

Interpret the result

Part (f)

step 1

Compare the adjusted

R

-square values from Table 1 and Table 2

step 2

Consider the significance of the predictors in each model

step 3

Conclude which model is preferred based on the comparison

Answer

[Insert final answer here]

Key Concept

Adjusted

R

-square

Explanation

Adjusted

R

-square accounts for the number of predictors in the model and provides a more accurate measure of model fit.

Key Concept

Sum of Squares Regression (SSR)

Explanation

SSR measures the variation explained by the regression model.

Key Concept

F-test for linear relationship

Explanation

The F-test determines if there is a significant linear relationship between the dependent and independent variables.

Key Concept

Partial F-test for dropping variables

Explanation

The partial F-test assesses whether a variable can be excluded from the model without significantly reducing the model's explanatory power.

Key Concept

Partial correlation

Explanation

Partial correlation measures the strength of the relationship between two variables while controlling for the effect of other variables.

Key Concept

Model comparison

Explanation

Model comparison involves evaluating the fit and significance of predictors to determine the preferred model.

Solution by Steps

step 1

To test the regression relation, we need to check the significance of the regression coefficients. We use the t-test for this purpose

step 2

The null hypothesis

H_0

is that the coefficient of temperature (

\beta_1

) is zero, i.e.,

H_0: \beta_1 = 0

step 3

The alternative hypothesis

H_a

is that the coefficient of temperature (

\beta_1

) is not zero, i.e.,

H_a: \beta_1 \neq 0

step 4

The t-statistic is calculated as

t = \frac{\text{Estimate}}{\text{Std. Error}} = \frac{-0.1537113}{0.0349408} \approx -4.40

step 5

We compare the t-statistic with the critical value from the t-distribution table at a given significance level (e.g.,

\alpha = 0.05

)

step 6

If the absolute value of the t-statistic is greater than the critical value, we reject the null hypothesis

step 7

Since

|t| \approx 4.40

is greater than the critical value (approximately 2.145 for 13 degrees of freedom), we reject

H_0

step 8

Therefore, there is a significant regression relation between temperature and yield

Answer

There is a significant regression relation between temperature and yield.

b) How much variation in

y

has been reduced when

x^{2}

is added in the model given that

x

is already in the model?

step 1

To determine the reduction in variation, we need to compare the

R^2

values of the models with and without

x^2

step 2

The

R^2

value for the model with

x

and

x^2

is given as

0.6732

step 3

The

R^2

value for the model with only

x

is not provided directly, but we can infer it from the given information

step 4

The reduction in variation is given by the difference in

R^2

values:

\Delta R^2 = R^2_{\text{with } x^2} - R^2_{\text{without } x^2}

step 5

Since the

R^2

value for the model with

x

and

x^2

is

0.6732

, and assuming the

R^2

value for the model with only

x

is lower, the reduction in variation is

0.6732 - R^2_{\text{without } x^2}

step 6

Without the exact

R^2

value for the model with only

x

, we cannot calculate the exact reduction, but we know it is significant

Answer

The variation in

y

has been significantly reduced when

x^2

is added to the model.

c) Justify whether the second-order polynomial regression model adequately fits the data.

step 1

To justify the adequacy of the model, we look at the

R^2

value and the significance of the regression coefficients

step 2

The

R^2

value of

0.6732

indicates that approximately 67.32% of the variation in yield is explained by the model

step 3

The significance of the regression coefficients can be checked using the t-tests

step 4

Both the linear term (temp) and the quadratic term (temp2) have significant coefficients, as indicated by their t-statistics

step 5

The model appears to fit the data well, as a high

R^2

value and significant coefficients suggest a good fit

Answer

The second-order polynomial regression model adequately fits the data.

d) Test whether the quadratic term can be dropped from the regression model.

step 1

To test whether the quadratic term can be dropped, we perform a t-test on the coefficient of the quadratic term (

\beta_2

)

step 2

The null hypothesis

H_0

is that the coefficient of the quadratic term (

\beta_2

) is zero, i.e.,

H_0: \beta_2 = 0

step 3

The alternative hypothesis

H_a

is that the coefficient of the quadratic term (

\beta_2

) is not zero, i.e.,

H_a: \beta_2 \neq 0

step 4

The t-statistic for the quadratic term is calculated as

t = \frac{\text{Estimate}}{\text{Std. Error}}

step 5

Since the standard error for the quadratic term is not provided, we cannot calculate the exact t-statistic

step 6

However, if the t-statistic is significant (i.e.,

|t|

is greater than the critical value), we reject

H_0

step 7

Given the context, it is likely that the quadratic term is significant, and thus, it should not be dropped

Answer

The quadratic term should not be dropped from the regression model.

Key Concept

Regression Analysis

Explanation

Regression analysis is used to determine the relationship between independent and dependent variables. In this case, it helps to understand how temperature affects yield.

Solution by Steps

step 1

The binomial model is appropriate for this distribution because each die roll is an independent Bernoulli trial with two possible outcomes: even or odd

step 2

The probability of getting an even number on a single die is

p = \frac{3}{6} = 0.5

step 3

The number of even scores in 4 rolls follows a binomial distribution

X \sim \text{Binomial}(n=4, p=0.5)

Part (b)

step 1

Calculate the expected frequencies for each number of even scores using the binomial formula:

P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

where

n=4

and

p=0.5

step 2

For

k=0

:

P(X=0) = \binom{4}{0} (0.5)^0 (0.5)^4 = 0.0625

. Expected frequency:

200 \times 0.0625 = 12.5

step 3

For

k=1

:

P(X=1) = \binom{4}{1} (0.5)^1 (0.5)^3 = 0.25

. Expected frequency:

200 \times 0.25 = 50

step 4

For

k=2

:

P(X=2) = \binom{4}{2} (0.5)^2 (0.5)^2 = 0.375

. Expected frequency:

200 \times 0.375 = 75

step 5

For

k=3

:

P(X=3) = \binom{4}{3} (0.5)^3 (0.5)^1 = 0.25

. Expected frequency:

200 \times 0.25 = 50

step 6

For

k=4

:

P(X=4) = \binom{4}{4} (0.5)^4 (0.5)^0 = 0.0625

. Expected frequency:

200 \times 0.0625 = 12.5

step 7

Calculate the

\chi^2

statistic:

\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}

where

O_i

are observed frequencies and

E_i

are expected frequencies

step 8

\chi^2 = \frac{(10-12.5)^2}{12.5} + \frac{(41-50)^2}{50} + \frac{(70-75)^2}{75} + \frac{(57-50)^2}{50} + \frac{(22-12.5)^2}{12.5} = 0.5 + 1.62 + 0.33 + 0.98 + 7.22 = 10.65

step 9

Compare the

\chi^2

statistic to the critical value from the

\chi^2

distribution table with

df = 4 - 1 = 3

at

\alpha = 0.05

. The critical value is 7.815

step 10

Since 10.65 > 7.815, we reject the null hypothesis that the sample comes from a binomial distribution

Answer

The binomial model might describe the distribution of

X

because each die roll is an independent Bernoulli trial with two possible outcomes: even or odd. The

\chi^2

-test at

\alpha=0.05

shows that the sample does not come from a binomial distribution.

Key Concept

Binomial Distribution and

\chi^2

-Test

Explanation

The binomial distribution is used to model the number of successes in a fixed number of independent Bernoulli trials. The

\chi^2

-test is used to compare observed frequencies with expected frequencies to determine if a sample comes from a specific distribution.

Solution by Steps

step 1

First, we need to calculate the expected frequencies for each category under the Poisson distribution with a mean of 4. The Poisson probability mass function is given by

P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}

where

\lambda = 4

step 2

Calculate the expected frequency for

x_i = 0

:

P(X = 0) = \frac{e^{-4} 4^0}{0!} = e^{-4} \approx 0.0183

. Multiply by the total number of observations (200):

0.0183 \times 200 \approx 3.66

step 3

Calculate the expected frequency for

x_i = 1

:

P(X = 1) = \frac{e^{-4} 4^1}{1!} = 4e^{-4} \approx 0.0733

. Multiply by 200:

0.0733 \times 200 \approx 14.66

step 4

Calculate the expected frequency for

x_i = 2

:

P(X = 2) = \frac{e^{-4} 4^2}{2!} = 8e^{-4} \approx 0.1465

. Multiply by 200:

0.1465 \times 200 \approx 29.30

step 5

Calculate the expected frequency for

x_i = 3

:

P(X = 3) = \frac{e^{-4} 4^3}{3!} = \frac{64e^{-4}}{6} \approx 0.1953

. Multiply by 200:

0.1953 \times 200 \approx 39.06

step 6

Calculate the expected frequency for

x_i = 4

:

P(X = 4) = \frac{e^{-4} 4^4}{4!} = \frac{256e^{-4}}{24} \approx 0.1953

. Multiply by 200:

0.1953 \times 200 \approx 39.06

step 7

Calculate the expected frequency for

x_i = 5

:

P(X = 5) = \frac{e^{-4} 4^5}{5!} = \frac{1024e^{-4}}{120} \approx 0.1562

. Multiply by 200:

0.1562 \times 200 \approx 31.24

step 8

Calculate the expected frequency for

x_i = 6

:

P(X = 6) = \frac{e^{-4} 4^6}{6!} = \frac{4096e^{-4}}{720} \approx 0.1041

. Multiply by 200:

0.1041 \times 200 \approx 20.82

step 9

Calculate the expected frequency for

x_i = 7

:

P(X = 7) = \frac{e^{-4} 4^7}{7!} = \frac{16384e^{-4}}{5040} \approx 0.0595

. Multiply by 200:

0.0595 \times 200 \approx 11.90

step 10

Calculate the expected frequency for

x_i = 8

:

P(X = 8) = \frac{e^{-4} 4^8}{8!} = \frac{65536e^{-4}}{40320} \approx 0.0297

. Multiply by 200:

0.0297 \times 200 \approx 5.94

step 11

Calculate the expected frequency for x_i > 8: This is the complement of the sum of probabilities from

x_i = 0

to

x_i = 8

. P(X > 8) = 1 - \sum_{i=0}^{8} P(X = i) \approx 1 - 0.9785 = 0.0215. Multiply by 200:

0.0215 \times 200 \approx 4.30

step 12

Now, we perform the

\chi^2

test. The formula for the

\chi^2

statistic is

\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}

where

O_i

is the observed frequency and

E_i

is the expected frequency

step 13

Calculate

\chi^2

for each category and sum them up:

\chi^2 = \frac{(5 - 3.66)^2}{3.66} + \frac{(12 - 14.66)^2}{14.66} + \frac{(31 - 29.30)^2}{29.30} + \frac{(40 - 39.06)^2}{39.06} + \frac{(38 - 39.06)^2}{39.06} + \frac{(29 - 31.24)^2}{31.24} + \frac{(22 - 20.82)^2}{20.82} + \frac{(14 - 11.90)^2}{11.90} + \frac{(5 - 5.94)^2}{5.94} + \frac{(4 - 4.30)^2}{4.30}

\chi^2 \approx 0.49 + 0.48 + 0.09 + 0.02 + 0.03 + 0.16 + 0.07 + 0.36 + 0.15 + 0.02 = 1.87

step 14

Compare the calculated

\chi^2

value with the critical value from the

\chi^2

distribution table with

df = 9 - 1 = 8

at

\alpha = 0.05

. The critical value is approximately 15.507. Since 1.87 < 15.507, we fail to reject the null hypothesis

Answer

The data fits the Poisson distribution with a mean of 4.

Key Concept

Chi-square test for goodness of fit

Explanation

The chi-square test compares the observed frequencies with the expected frequencies under a specified distribution to determine if the observed data fits the distribution.

Solution by Steps

step 1

To test the hypothesis that the duration of the effect is exponentially distributed, we first need to calculate the expected frequencies for each interval. The exponential distribution has the probability density function

f(x) = \lambda e^{-\lambda x}

, where

\lambda

is the rate parameter

step 2

We estimate

\lambda

using the sample mean. The sample mean

\bar{x}

can be calculated as follows:

\bar{x} = \frac{\sum (x_i \cdot f_i)}{N}

where

x_i

is the midpoint of each interval,

f_i

is the frequency, and

N

is the total number of observations

step 3

Calculate the midpoints of each interval:

\text{Midpoints} = \{1.5, 4.5, 7.5, 10.5, 15, 21, 30\}

step 4

Calculate the sample mean

\bar{x}

:

\bar{x} = \frac{(1.5 \cdot 40) + (4.5 \cdot 31) + (7.5 \cdot 31) + (10.5 \cdot 22) + (15 \cdot 23) + (21 \cdot 22) + (30 \cdot 31)}{200}

\bar{x} = \frac{60 + 139.5 + 232.5 + 231 + 345 + 462 + 930}{200} = \frac{2400}{200} = 12

step 5

Estimate

\lambda

:

\lambda = \frac{1}{\bar{x}} = \frac{1}{12}

step 6

Calculate the expected frequencies for each interval using the cumulative distribution function (CDF) of the exponential distribution:

F(x) = 1 - e^{-\lambda x}

step 7

Calculate the expected frequencies for each interval:

\text{Expected frequency for } 0-3 = 200 \cdot (F(3) - F(0)) = 200 \cdot (1 - e^{-\frac{3}{12}} - 0) = 200 \cdot (1 - e^{-0.25})

\text{Expected frequency for } 3-6 = 200 \cdot (F(6) - F(3)) = 200 \cdot (1 - e^{-\frac{6}{12}} - (1 - e^{-\frac{3}{12}})) = 200 \cdot (e^{-0.25} - e^{-0.5})

\text{Expected frequency for } 6-9 = 200 \cdot (F(9) - F(6)) = 200 \cdot (e^{-0.5} - e^{-0.75})

\text{Expected frequency for } 9-12 = 200 \cdot (F(12) - F(9)) = 200 \cdot (e^{-0.75} - e^{-1})

\text{Expected frequency for } 12-18 = 200 \cdot (F(18) - F(12)) = 200 \cdot (e^{-1} - e^{-1.5})

\text{Expected frequency for } 18-24 = 200 \cdot (F(24) - F(18)) = 200 \cdot (e^{-1.5} - e^{-2})

\text{Expected frequency for } 24-36 = 200 \cdot (F(36) - F(24)) = 200 \cdot (e^{-2} - e^{-3})

step 8

Perform a chi-square goodness-of-fit test:

\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}

where

O_i

are the observed frequencies and

E_i

are the expected frequencies. Compare the calculated

\chi^2

value with the critical value from the chi-square distribution table with appropriate degrees of freedom to determine if the null hypothesis can be rejected

Answer

The hypothesis that the duration of the effect is exponentially distributed can be tested using the chi-square goodness-of-fit test.

Key Concept

Exponential Distribution and Chi-Square Test

Explanation

The exponential distribution is characterized by its rate parameter

\lambda

, which can be estimated using the sample mean. The chi-square goodness-of-fit test compares observed frequencies with expected frequencies to determine if the data follows the hypothesized distribution.

Solution by Steps

step 1

Identify the Poisson distribution for one counter. Given that each counter serves an average of 10 customers in 30 minutes, the parameter

\lambda

for the Poisson distribution is 10

step 2

Write the probability distribution function (PDF) for the number of customers served by one counter in 30 minutes. The PDF of a Poisson distribution with parameter

\lambda

is given by:

P(X = x) = \frac{\lambda^x e^{-\lambda}}{x!}

Substituting

\lambda = 10

:

P(X = x) = \frac{10^x e^{-10}}{x!}

step 3

For part (b), since the counters are independent and identically distributed, the total number of customers served by all 5 counters in 30 minutes is the sum of 5 independent Poisson random variables, each with

\lambda = 10

step 4

The sum of independent Poisson random variables is also a Poisson random variable with parameter equal to the sum of the individual parameters. Therefore, the total number of customers served by all 5 counters follows a Poisson distribution with parameter

5 \times 10 = 50

step 5

Write the probability distribution function for the total number of customers served by all 5 counters:

P(Y = y) = \frac{50^y e^{-50}}{y!}

where

Y

is the total number of customers served by all 5 counters in 30 minutes

step 6

State the mean and variance of the total number of customers served by all 5 counters. For a Poisson distribution with parameter

\lambda

, the mean and variance are both equal to

\lambda

. Therefore, for

\lambda = 50

:

\text{Mean} = 50

\text{Variance} = 50

Answer

The probability distribution function for the number of customers served by one counter in 30 minutes is:

P(X = x) = \frac{10^x e^{-10}}{x!}

The probability distribution function for the total number of customers served by all 5 counters in 30 minutes is:

P(Y = y) = \frac{50^y e^{-50}}{y!}

The mean and variance of the total number of customers served by all 5 counters are both 50.

Key Concept

Poisson Distribution

Explanation

The Poisson distribution is used to model the number of events occurring within a fixed interval of time or space. The sum of independent Poisson random variables is also a Poisson random variable with a parameter equal to the sum of the individual parameters.

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