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DIA
8 days ago
11

The following estimated regression model was developed relating yearly income (y in $1000s) of 30 individuals with their age (x1

) and their gender (x2) (0 if male and 1 if female).
ŷ = 30 + 0.7x1 + 3x2

Also provided are SST = 1200 and SSE = 384.
The multiple coefficient of determination is _____.

a. .50
b. .42
c. .68
d. .32
Mathematics
1 answer:
zzz [12.3K]8 days ago
4 0
Thus, the most suitable answer is b..42.Step-by-step reasoning:Prior concepts include an explanation of Analysis of variance (ANOVA), which is employed to assess variances among group averages within a sample. The sum of squares constitutes the total squared variation where variation is defined as the disparity between each value and the grand mean. The correlation coefficient evaluates the strength of the correlation between two variable movements, noted as r, with values bounded between -1 and 1. In conducting multiple regression analysis, we seek to ascertain the relationship among multiple independent (predictor) variables and a dependent (criterion) variable.Solution:Assuming the presence of k independent variables and j=1,\dots,j individuals, we can articulate various formulas of variation: We also possess a characteristic identified as SST=SS_{regression}+SS_{error}. The model's degrees of freedom in this circumstance is represented by df_{model}=df_{regression}=k=2, with k =2 indicative of the variable count. The error's degrees of freedom is articulated by df_{error}=N-k-1=30-2-1=27. The coefficient of determination in multiple regression is illustrated as: thus, the answer is b..42.
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A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of awriting
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a) Null hypothesis:\mu \geq 10

Alternative hypothesis:\mu < 10

b) p_v =P(t_{17}

Given that the p-value is lower than the significance threshold in this situation, we have sufficient grounds to reject the null hypothesis.

c) p_v =P(t_{17}

In this case, since the p-value exceeds the significance threshold, we have adequate evidence to FAIL to reject the null hypothesis.

d) p_v =P(t_{17}

Here again, with the p-value being less than the significance level, we can reject the null hypothesis.

Step-by-step explanation:

1) Provided data and references

\bar X represents the average of the samples

s denotes the standard deviation of the samples

n=18 indicates the number of samples

\mu_o =10 is the value we are examining

\alpha defines the significance level for the test.

t represents the specific statistic of interest

p_v indicates the p-value relevant to the test (the variable of concern)

Define the null and alternative hypotheses.

To assess if the true mean is at least 10 hours, we must set up a hypothesis:

Part a

Null hypothesis:\mu \geq 10

Alternative hypothesis:\mu < 10

If we consider the sample size being less than 30 and the population deviation unknown, it’s more appropriate to use a t-test to compare the actual mean with the reference value, calculated as:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)

Part b

In this scenario t=-2.3, \alpha=0.05

Initially, we need to calculate the degrees of freedom df=n-1=18-1=17

Since this is a left-tailed test, the p-value is determined by:

p_v =P(t_{17}

In this instance, with the p-value being less than the significance level, we have sufficient evidence to reject the null hypothesis.

Part c

For this situation t=-1.8, \alpha=0.01

We need to find the degrees of freedom df=n-1=18-1=17

For the left-tailed test, the p-value is given by:

p_v =P(t_{17}

In this case, since the p-value is above the significance level, we have enough grounds to FAIL to reject the null hypothesis.

Part d

For this case t=-3.6, \alpha=0.05

Firstly, we find the degrees of freedom df=n-1=18-1=17

Since we are conducting a left-tailed test, the p-value is calculated as:

p_v =P(t_{17}

Here, with the p-value being lower than the significance threshold, we can reject the null hypothesis.

5 0
1 month ago
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