Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

what is the a d polnn) Based on the simple egession moel above, of 30 mm Hg? b,

ID: 3326619 • Letter: W

Question

what is the a d polnn) Based on the simple egession moel above, of 30 mm Hg? b, (4 point) Calulate the approximate (assuming mean for an atmospherie pressure is 30 mmHg) 95% prediction interval for your estimate in part (a) above e) (4 points) Calculate the 95% confidence interval for by, the coefficient for pressure the variable atmospheric e) (4 points) Perform the appropriate hypothesis test to determine whether atmospheric pressure is a significant predictor of ozone concentration levels. Be sure to list the hypotheses in terms of parameters, the test statistic and degrees of freedom if applicable, the p-value, and whether or not you reject the null hypothesis.

Explanation / Answer

a)

Based on the result provided,

The estimated ozone concentration level on a day

= 1095.183 + 7.571 * atmospheric pressure

Hence when the atmospheric pressure is 30 mmHg,

The estimated ozone concentration is equal to

1095.183 + 7.571 * 30

= 1322.313 in parts per 100 Million

b)

The 95% prediction interval of the ozone concentration when the atmospheric pressure is 30 mmHg is:

The lower 95 prediction interval is

= (1095.183 - 67.041) + (7.571 - 2.076) * 30

= 1192.992

The upper 95 prediction interval is

= (1095.183 + 67.041) + (7.571 + 2.076) * 30

= 1451.654

c)

95 confidence interval for the variable b1, the coefficient of atmospheric pressure is:

Lower 95 confidence interval is

= 7.571 - 2.076

= 5.495

Upper 95 confidence interval

= 7.571 + 2.076

= 9.647

d)

To determine whether the atmospheric pressure is a significant prediction of ozone concentration

Null hypothesis is: b1 (the coefficient of the variable atmospheric pressure) = 0

Alternate hypothesis is: b1 not equal to 0

The t statistic is = 7.571 / 2.076 = 3.647

The degrees of freedom for this t statistic = 1

So, the p-value of the hypothesis test is:

= 2 * P(T > 3.647 | T ~ t(1))

= 2 * (1 - pt(3.647,1))

= 0.1703

Which is greater than level of significance.

So we fail to reject the null hypothesis that b1 = 0

Hence we conclude that:

The atmospheric pressure is not a significant predictor of ozone concentration.