opatovaL

2020-12-12

Ozone The Environmental Protection Agency is examin-
ing the relationship between the ozone level (in parts per
million) and the population (in millions) of U.S. cities.

Part of the regression analysis is shown.

Dependent variable is: Ozone

R-squared$=84.4\mathrm{\%}$

$s=5.454$
with $16-2=14$ df

Variable Coefficient SE(Coeff)

Intercept 18.892 2.395

Pop 6.650 1.910

a) We suspect that the greater the population of a city, the higher its ozone level. Is the relationship significant? Assuming the conditions for inference are satisfied, test an appropriate hypothesis and state your conclu- sion in context.

b) Do you think that the population of a city is a useful predictor of ozone level? Use the values of both R2 and s in your explanation.

Part of the regression analysis is shown.

Dependent variable is: Ozone

R-squared

Variable Coefficient SE(Coeff)

Intercept 18.892 2.395

Pop 6.650 1.910

a) We suspect that the greater the population of a city, the higher its ozone level. Is the relationship significant? Assuming the conditions for inference are satisfied, test an appropriate hypothesis and state your conclu- sion in context.

b) Do you think that the population of a city is a useful predictor of ozone level? Use the values of both R2 and s in your explanation.

insonsipthinye

Skilled2020-12-13Added 83 answers

a) The null and alternative hypotheses are,

$H0:\beta 1=0$

That is, there is no linear relationship between population and ozone level.

$H1:\beta 1\ne 0$

That is, there is a linear relationship between population and ozone level.

Degrees of freedom:

The degrees of the freedom is,

$n-2=16-2=14$

The formula for finding the test statistic is,

$t=\frac{{b}_{1}-{\beta}_{1}}{SE({b}_{1})}$

Subsitute 6.650 for b1, for beta1 and 1.910 for SE(b1).

$t=\frac{{b}_{1}-{\beta}_{1}}{SE({b}_{1})}$

$\frac{6.650-0}{1.910}$

$=3.48$

Thus, the test statistic is 3.48.

Computation of P-value:

The P-value of t-distribution at 14 degrees of freedom can be obtained using the excel formula$\u201c=T.DIST.2T(3.48,14)\u201d$

. Thus, the P-value is 0.0037.

It means that there is association in the data is unlikely to occur by chance. Thus, we reject the null hypothesis.

Hence, there is sufficient evidence that there is a positive linear relationship between ozone level and population. Therefore the conclusion is, cities with larger populations tend to have higher ozone levels,

b) From output, it can be observed the R scored is 84.4% and s value is 5.454. Therefore, the population explains about 84% of the variability in ozone level and s is over 5 parts per million. Thus, the population is a good predictor of ozone level.

Yes, the population of a city is a useful predictor of ozone level.

That is, there is no linear relationship between population and ozone level.

That is, there is a linear relationship between population and ozone level.

Degrees of freedom:

The degrees of the freedom is,

The formula for finding the test statistic is,

Subsitute 6.650 for b1, for beta1 and 1.910 for SE(b1).

Thus, the test statistic is 3.48.

Computation of P-value:

The P-value of t-distribution at 14 degrees of freedom can be obtained using the excel formula

. Thus, the P-value is 0.0037.

It means that there is association in the data is unlikely to occur by chance. Thus, we reject the null hypothesis.

Hence, there is sufficient evidence that there is a positive linear relationship between ozone level and population. Therefore the conclusion is, cities with larger populations tend to have higher ozone levels,

b) From output, it can be observed the R scored is 84.4% and s value is 5.454. Therefore, the population explains about 84% of the variability in ozone level and s is over 5 parts per million. Thus, the population is a good predictor of ozone level.

Yes, the population of a city is a useful predictor of ozone level.

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