Question

Listed below are numbers of Internet users per 100 people and numbers of scientific award winners per 10 million people for different countries. Construct a scatterplot, find the value of the linear correlation coefficient r, and find the P-value of r. Determine whether there is sufficient evidence to support a claim of linear correlation between the two variables. Use a significance level of a = 0.01. Internet Users 79.1 80.5 57.2 68.8 77.9 37.6 Award Winners 5.3 9.2 1.7 10.9 0.1 3.4 Construct a scatterplot. Choose the correct graph below. OA B. OD. 12. 12 12 BOL Q Award Winners . 0+ 30 90 30 90 0+ 30 90 30 90 Internet Users Internet Users Internet Users Internet Users The linear correlation coefficient ris | (Round to three decimal places as needed.)

The test statistic is...............(Round to two decimal places as​ needed.)

The​ P-value is.......................(Round to three decimal places as​ needed.)

The test statistic t is .................. ​(Round to three decimal places as​ needed.)

The​ P-value is.............................(Round to three decimal places as​ needed.)

The​ P-value for this hypothesis test is

0.2300.230.

​(Round to three decimal places as​ needed.)

Answer 1

Q1) The value of the correlation coefficient is $r\approx 0.782$

The value of the test static is $t={r\sqrt{n-2}\over \sqrt{1-r^2}}= {0.782\sqrt{6-2}\over \sqrt{1-(0.782)^2}} \approx 2.51$

The test is two tailed test and number of degrees of freedom is 6-2=4

So P-value is $\Pr(|t|>2.51)=0.066$

As P-value is 0.066>0.01

We fail to reject the null hypothesis

There is no relation between the two

Q2)

The value of the correlation coefficient is $r\approx 0.862$

The value of the test static is

ryn - 2 ta V1 - 72 0.8629 – 2 11 - (0.862)2 4.499

The test is two tailed test and number of degrees of freedom is 9-2=7

So P-value is $\Pr(|t|>4.499)=0.003$

As P-value is 0.003<0.05

We reject the null hypothesis

Q3) The sample proportion is

$\hat{p }= {55\over 239}\approx 0.23$

Value of test static is

$z={\hat{p }-p\over \sqrt{p(1-p)\over n}}={0.23-0.20 \over \sqrt{0.20(1-0.20)\over 239}} \approx 1.16$

The P-value is $\Pr(z>1.16)=0.1230$