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kari74 [83]
3 years ago
13

Statistics students in Oxnard College sampled 10 textbooks in the Condor bookstore, and recorded number of pages in each textboo

k and its cost. The bivariate data is shown below, Number of Pages ( x ) Cost( y ) 526 52.08 625 59 589 56.12 409 25.72 489 34.12 500 53 906 78.48 251 26.08 595 50.6 719 68.52 A student calculates a linear model y = x + . (Please show your answers to two decimal places) Use the model above to estimate the cost when number of pages is 563 Cost = $ (Please show your answer to 2 decimal places.)
Mathematics
1 answer:
jeka943 years ago
4 0

Answer:

y = -0.85 + 0.09x; $49.82

Step-by-step explanation:

1. Calculate Σx, Σy, Σxy, and Σx²  

The calculation is tedious but not difficult.

\begin{array}{rrrr}\mathbf{x} & \mathbf{y}  & \mathbf{xy} & \mathbf{x^{2}}\\526 & 52.08 & 27394.08 & 276676\\625& 59.00 & 36875.00 &390625\\589 & 56.12 & 33054.68 & 346921\\409 & 25.72 & 10519.48 & 167281\\489 & 34.12& 16684.68 & 293121\\500 & 53.00 & 26500.00 &250000\\906 & 76.48 & 71102.88 & 820836\\251 &26.08 & 6546.08 & 63001\\595 & 50.60 & 30107.00 & 354025\\719 & 68.52 & 49265.88 & 516961\\\mathbf{5609} & \mathbf{503.72} &\mathbf{308049.76} & \mathbf{3425447}\\\end{array}

2. Calculate the coefficients in the regression equation

a = \dfrac{\sum y \sum x^{2} - \sum x \sum xy}{n\sum x^{2}- \left (\sum x\right )^{2}} =  \dfrac{503.7 \times 3425447 - 5609 \times 308049.76}{10 \times 3425447- 5609^{2}}\\\\= \dfrac{1725466163 - 1727851103.84}{34254470 - 31460881} = -\dfrac{2384941}{2793589}= \mathbf{-0.8537}

b = \dfrac{n\sumx y  - \sum x \sumxy}{n\sum x^{2}- \left (\sum x\right )^{2}} = \dfrac{3080498 - 2825365.48}{2793589} = \dfrac{255132}{2793589} = \mathbf{0.09133}

To two decimal places, the regression equation is

y = -0.85 + 0.09x

3. Prediction

If x = 563,

y = -0.85 + 0.09x = -0.85 + 0.09 × 563 = -0.85  + 50.67 = $49.82

(If we don't  round the regression equation to two decimal places, the predicted value is $50.56.)

 

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a)z=4.658

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b) Using the significance level assumed \alpha=0.05 we see that p_v so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life is  higher than 50%.

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We are confident that about 54% to 59% of all Americans think the Civil War is relevant.

Step-by-step explanation:

I )Part a

1) Data given and notation

n=1507 represent the random sample taken  

X represent the Americans who thinks that the Civil War is still relevant to American politics and political life

\hat p estimated proportion of Americans who thinks that the Civil War is still relevant to American politics and political life in the sample

p_o=0.5 is the value that we want to test since the problem says majority    

\alpha=0.05 represent the significance level (no given, but is assumed)  

z would represent the statistic (variable of interest)  

p_v represent the p value (variable of interest)  

p= population proportion of Americans who thinks that the Civil War is still relevant to American politics and political life

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life exceeds 50%(Majority). :  

Null Hypothesis: p \leq 0.5

Alternative Hypothesis: p >0.5

We assume that the proportion follows a normal distribution.  

This is a one tail upper test for the proportion of  union membership.

The One-Sample Proportion Test is "used to assess whether a population proportion \hat p is significantly (different,higher or less) from a hypothesized value p_o".

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

np_o =1507*0.5=753.5>10

n(1-p_o)=1507*(1-0.5)=753.5>10

3) Calculate the statistic  

The statistic is calculated with the following formula:

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

On this case the value of p_o=0.5 is the value that we are testing and n = 1507.

z=\frac{0.56 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1507}}}=4.658

The p value for the test would be:

p_v =P(z>4.65)=1-P(z  

II) Part b

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

Based on the alternative hypothesis the p value would be given by:

p_v =P(z>4.65)=1-P(z  

Using the significance level assumed \alpha=0.05 we see that p_v so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life is  higher than 50%.

III) Part c

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 90% confidence interval the value of \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

0.56 - 2.58 \sqrt{\frac{0.56(1-0.56)}{1507}}=0.527

0.56 + 2.58 \sqrt{\frac{0.56(1-0.56)}{1507}}=0.593

And the 90% confidence interval would be given (0.527;0.593).

We are confident that about 54% to 59% of all Americans think the Civil War is relevant.

And this result agrees with the result of part b, since the interval not contains the value of 0.5 we can conclude that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life it's higher than 0.5 at 90% of confidence.

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