MATHEMATICS COLLEGE

The table below gives the list price and the number of bids received for five randomly selected items sold through online auctions. Using this data, consider the equation of the regression line, yˆ=b0+b1x, for predicting the number of bids an item will receive based on the list price. Keep in mind, the correlation coefficient may or may not be statistically significant for the data given. Remember, in practice, it would not be appropriate to use the regression line to make a prediction if the correlation coefficient is not statistically significant. Price in Dollars 25 33 34 45 48
Number of Bids 2 3 4 5 7

Step 1 of 6: Find the estimated slope. Round your answer to three decimal places.
Step 2 of 6: Find the estimated y-intercept. Round your answer to three decimal places.Step 3 of 6: Find the estimated value of y when x = 34. Round your answer to three decimal places.Step 4 of 6: Determine the value of the dependent variable yˆ at x = 0.Step 5 of 6: Substitute the values you found in steps 1 and 2 into the equation for the regression line to find the estimated linear model. According to this model, if the value of the independent variable is increased by one unit, then find the change in the dependent variable yˆ.Step 6 of 6: Find the value of the coefficient of determination.

Answers

Answer 1

Answer:

Answer:

1) b1=5.831

2) b0=12.510

3) y(34)=210.764

4) y(0)=12.510

5) y=12.510+5.831x

6) R^2=0.85

Step-by-step explanation:

We have the linear regression model $y=b_0+b_1 x$ .

We start by calculating the all the parameters needed to define the model:

- Mean of x:

$\bar x=(1)/(5)\sum_(i=1)^(5)(2+3+4+5+7)=(21)/(5)=4.2$

- Uncorrected standard deviation of x:

$s_x=\sqrt{(1)/(n)\sum_(i=1)^(5)(x_i-\bar x)^2}\n\n\ns_x=\sqrt{(1)/(5)\cdot [(2-4.2)^2+(3-4.2)^2+(4-4.2)^2+(5-4.2)^2+(7-4.2)^2]}\n\n\n s_x=\sqrt{(1)/(5)\cdot [(4.84)+(1.44)+(0.04)+(0.64)+(7.84)]}\n\n\n s_x=\sqrt{(14.8)/(5)}=√(2.96)\n\n\ns_x=1.72$

- Mean of y:

$\bar y=(1)/(5)\sum_(i=1)^(5)(25+33+34+45+48)=(185)/(5)=37$

- Standard deviation of y:

$s_y=\sqrt{(1)/(n)\sum_(i=1)^(5)(y_i-\bar y)^2}\n\n\ns_y=\sqrt{(1)/(5)\cdot [(25-37)^2+(33-37)^2+(34-37)^2+(45-37)^2+(48-37)^2]}\n\n\n s_y=\sqrt{(1)/(5)\cdot [(144)+(16)+(9)+(64)+(121)]}\n\n\n s_y=\sqrt{(354)/(5)}=√(70.8)\n\n\ns_y=8.414$

- Sample correlation coefficient

$r_(xy)=\sum_(i=1)^5((x_i-\bar x)(y_i-\bar y))/((n-1)s_xs_y)\n\n\nr_(xy)=((2-4.2)(25-37)+(3-4.2)(33-37)+...+(7-4.2)(48-37))/(4\cdot 1.72\cdot 8.414)\n\n\nr_(xy)=(69)/(57.888)=1.192$

Step 1

The slope b1 can be calculated as:

$b_1=r_(xy)(s_y)/(s_x)=1.192\cdot(8.414)/(1.72)=5.831$

Step 2

The y-intercept b0 can now be calculated as:

$b_o=\bar y-b_1\bar x=37-5.831\cdot 4.2=37-24.490=12.510$

Step 3

The estimated value of y when x=34 is:

$y(34)=12.510+5.831\cdot(34)=12.510+198.254=210.764$

Step 4

At x=0, the estimated y takes the value of the y-intercept, by definition.

$y(0)=12.510+5.831\cdot(0)=12.510+0=12.510$

Step 5

The linear model becomes

$y=12.510+5.831x$

Step 6

The coefficient of determination can be calculated as:

$R^2=1-(SS_(res))/(SS_(tot))=1-(\sum(y_i-f_i))/(ns_y^2)\n\n\n\sum(y_i-f_i)=(25-24.17)^2+(33-30)^2+(34-35.83)^2+(45-41.67)^2+(48-53.33)^2\n\n\sum(y_i-f_i)=0.69+ 8.98+ 3.36+ 11.12+ 28.38=52.53\n\n\n ns_y^2=5\cdot 8.414^2=353.98\n\n\nR^2=1-(52.53)/(353.98)=1-0.15=0.85$

This should be Correct-

Answer:

True

Explanation:

A statistic is a characteristic of a sample, a portion of the target population. A parameter is a fixed, unknown numerical value, while the statistic is a known number and a variable which depends on the portion of the population.

Parameter Definition: a quantity or statistical measure that, for a given population, is fixed and that is used as the value of a variable in some general distribution or frequency function to make it descriptive of that.

Population: The mean and variance of a population are population parameters.

Statistic Definiton: A statistic or sample statistic is any quantity computed from values in a sample that is used for a statistical purpose. Statistical purposes include estimating a population parameter, describing a sample, or evaluating a hypothesis. The average of sample values is a statistic.

The answer is: true

Hope this helps! Happy holidays;)

Will mark brainest help please !!!

Answers

8) red and 1, red and 2, red and 3, blue and 1, blue and 2, blue and 3

So answer is 6

9) 0.5*1/3 = 0.17

10) 0.5*2/3 = 0.33

The percent of fat calories that a person consumes each day is normally distributed with a mean of 35 and a standard deviation of 10. Suppose that 16 individuals are randomly chosen. Let X = average percent of fat calories. (a) For the group of 16 individuals, find the probability that the average percent of fat calories consumed is more than thirty-seven. (Round your answer to four decimal places.) b) Find the first quartile for the average percent of fat calories. (Round your answer to two decimal places.) percent of fat calories

Answers

Answer:

a) 0.2119 = 21.19% probability that the average percent of fat calories consumed is more than thirty-seven.

b) The first quartile for the average percent of fat calories is 33.31

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = (X - \mu)/(\sigma)$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = (\sigma)/(√(n))$

In this problem, we have that:

$\mu = 35, \sigma = 10, n = 16, s = (10)/(√(16)) = 2.5$

(a) For the group of 16 individuals, find the probability that the average percent of fat calories consumed is more than thirty-seven. (Round your answer to four decimal places.)

This is the 1 subtracted by the pvalue of Z when X = 37. So

$Z = (X - \mu)/(\sigma)$

By the Central Limit Theorem

$Z = (X - \mu)/(s)$