g. Does it appear that a line is the best way

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size (inches)

9  11  20 27 32 35 40 48 53 60

sale price (s)

147 186 197 297 447 1177 1228 1463 1650 178

e.Find the correlation coefficient. Is it significant?
f. Find the estimated sale price for a 32 and 52 inch table.

g. Does it appear that a line is the best way to fit the data? Why or why not? h. What is the slope of the least-squares (best-fit) line? Interpret the slope.

 

Expert Solution
Step 1

Here the explanatory variable is x such that size and the predicted variable is y such that price.

The data given for size (inches) and sale price (s) is listed below

x y
9 147
11 186
20 197
27 297
32 447
35 1177
40 1228
48 1463
53 1650
60 178

 

It is required to test whether correlation coefficient is significant.

Step 2: f)

First obtaining the least square and the correlation coefficient.

It can be solved using MS-Excel.

Steps for obtaining least square and the correlation coefficient the  using MS-Excel are as follows

  • Enter the data in MS-Excel
  • Click on the Data toolbar >> Select the Data Analysis >> Regression
  • Select the entire X column data as an input  X range. (Check labels if included)
  • Select the entire Y column data as an input  Y range. (Check labels if included)
  • Press ok

The output obtained is as follows

Statistics homework question answer, step 2, image 1

From the above output

The correlation coefficeint is r = 0.5827

As given,

The correlation coefficient (r) between the amounts of    is

The sample size is n = 10

 

Now we want to test the significance of the correlation coefficient at 5% level of significance.

The most appropriate test is the significance of the correlation coefficient test.

Let ρ  be the population correlation coefficient between the variables ().

 

To test the significance the null hypothesis is given as

H0:ρ=0

That is there no is a significant correlation between the variables.

 

Vs the alternative hypothesis is given as

H1:ρ0

That is there is a significant correlation between the variables.

 

Now obtaining the test statistic.

The test statistic is given as

t=r1-r2n-2

t=0.58271-0.5827210-2

=

 

The degree of freedom is given as

Df = n -2

= 8

 

Now obtaining the p-value.

Since the hypothesis is two-tailed the p-value for test statistic t = 2.027993  with df = 8  is given as

 

 =   0.0771                         


                                    Using the MS-EXCEL command (T.DIST.2T(2.027993,8 ))

Decision rule: -

At 5% level of significance

Reject the null hypothesis if p-value < 0.05

Accept otherwise.

Since p-value =  0.0771    >  0.05

We failed to  reject the null hypothesis.

Conclusion: -

At 5% level of significance, there is no sufficient evidence to conclude that there is significant correlation between the variables.

 

 

 

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