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Over [174]
3 years ago
10

The admissions officer for Clearwater College developed the following estimated regression equation relating the final college G

PA to the student's SAT mathematics score and high-school GPA. where high-school grade point average SAT mathematics score final college grade point average Round your answers to 4 decimal places.
a. Complete the missing entries in this Excel Regression tool output. Enter negative values as negative numbers. ANOVA df SS MS F Significance F Regression 2 .8811 52.4464 Residual 7 .1179 .0168 Total Coefficients Standard Error t Stat P-value Intercept X1 X2
b. Using , test for overall significance.
c. Did the estimated regression equation provide a good fit to the data
Mathematics
1 answer:
Strike441 [17]3 years ago
5 0

Answer:

A. Complete the missing entries in this Excel Regression tool output. Enter negative values as negative numbers. ANOVA df SS MS F Significance F Regression 2 .8811 52.4464 Residual 7 .1179 .0168 Total Coefficients Standard Error t Stat P-value Intercept X1 X2.

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F (x)<br> 15x + 13 Which is the average rate of change over for the interval 0 &lt; x &lt; 10?
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3 years ago
A random sample of 150 men found that 88 of the men excercise regularly, while a random sample 200 women found that 130 of the w
melamori03 [73]

Answer:

The hypothesis is:

<em>H₀</em>: p_{X}-p_{Y}=0.

<em>Hₐ</em>: p_{X}-p_{Y}.

Step-by-step explanation:

Let <em>X</em> = number of men who exercise regularly and <em>Y</em> = number of women who exercise regularly.

The information provided is:

n_{X}=150\\X=88\\n_{Y}=200\\Y=130

Compute the sample proportion of men and women who exercise regularly as follows:

\hat p_{X}=\frac{X}{n_{X}}=\frac{88}{150}=0.587

\hat p_{Y}=\frac{Y}{n_{Y}}=\frac{130}{200}=0.65

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 150 and \hat p_{X}=0.587.

The random variable <em>Y</em> also follows a Binomial distribution with parameters <em>n</em> = 200 and \hat p_{Y}=0.65.

According to the Central limit theorem, if from an unknown population large samples of sizes <em>n</em> > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

\hat p=p

The standard deviation of this sampling distribution of sample proportion is:

\sigma_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}

So, the sampling distribution of the proportion of men and women who exercise regularly follows a Normal distribution.

A two proportion <em>z</em>-test cab be performed to determine whether the proportion of women is more than men who exercise regularly.

The hypothesis for this test cab be defined as:

<em>H₀</em>: The proportion of women is same as men who exercise regularly, i.e. p_{X}-p_{Y}=0.

<em>Hₐ</em>: The proportion of women is more than men who exercise regularly, i.e. p_{X}-p_{Y}.

6 0
3 years ago
Read 2 more answers
Is 272 oz greater than 20 lb
Karolina [17]
We know that 1 lb = 16 oz
So lets convert the 20 lb to oz and then compare the two quantities in consistent units.
If 1 lb = 16 oz then 20 lb are equal to 20*16 oz, that is
20 lb = 20*16 oz = 320 oz
therefore,
320 oz are greater than 272 oz, or,
20 lb are greater that 272 oz
3 0
3 years ago
+12&lt;16<br> Please with steps :)
mixas84 [53]

Answer:

0,1,2,3

Step-by-step explanation:

6 0
2 years ago
Suppose you like to keep a jar of change on your desk. Currently, the jar contains the following:
antoniya [11.8K]

Answer:

The probability is 435/6,806

Step-by-step explanation:

Firstly, the total number of coins in the jar is

13 + 15 + 29 + 26 = 83

The probability of selecting a dime is 15/83

Now we want to grab a nickel

Since we already selected a dime and we did not replace it, the number of total coins in the jar will be 83-1 = 82

So the probability of selecting a nickel this time is

29/82

So the total probability of the event in the described order is;

15/83 * 29/82

= 435/6,806

4 0
2 years ago
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