Question

In: Statistics and Probability

The ozone dataset gives a sample of ozone measurements (in partial pressures) taken over the South...

The ozone dataset gives a sample of ozone measurements (in partial pressures) taken over the South Pole on September 18, 1997 at altitudes above 10km.

mPa:

1.828, 2.652, 3.307, 3.855, 4.021, 4.173, 4.316, 4.951, 4.787, 4.554, 4.333, 4.039, 4.48, 4.739, 4.791, 5.213, 5.75, 5.79, 5.656, 5.464, 5.092, 3.135, 2.754, 3.14, 5.213, 5.831, 5.736, 5.333, 4.797, 6.704, 6.493, 5.746, 6.31, 6.53, 6.63, 6.071, 5.706, 4.951, 4.392, 4.619, 5.029, 5.302, 5.151, 3.474, 3.285, 3.232, 3.4, 3.503, 3.649, 3.828, 4.235, 4.781, 5.096, 5.262, 5.411, 5.439, 5.08, 4.719, 4.519

1. Compute the five-number summary and sketch the boxplot. Identify any outliers.

2. Compute the mean and standard deviation of the sample.

3. Construct the probability plot of the data. A hypothesis test could be used to compare the mean ozone level on September 18, 1997 to a specified baseline level.

4. Are the assumptions for an hypothesis test of the mean reasonably satisfied?

5. Test to see if the mean ozone amount on September 18 was below 5 mPa, at the 5% level of significance.

Year

106 × km2

1979

2.23

1980

1.88

1981

1.70

1982

3.77

1983

6.24

1984

8.66

1985

12.57

1986

9.58

1987

18.18

1988

8.75

1989

17.75

1990

17.86

1991

18.13

1992

21.28

1993

22.81

1994

22.82

6. Construct a line graph of the data in the table. Note any trends that you see.

7. Does it make sense to apply inferential methods (of the type we have studied) to the mean size of the ozone hole over time? Explain.

In addition to plotting the ozone concentration versus time, create a linear regression model for it and interpret the slope and comment on how well the model fits the data using the coefficient of variation.

(IF POSSIBLE INCLUDE R CODES WITH THE ANSWERS FOR EACH QUESTION)

Solutions

Expert Solution

Solution-1:

R code is

ozone <- c(1.828, 2.652, 3.307, 3.855, 4.021, 4.173, 4.316, 4.951, 4.787, 4.554, 4.333, 4.039, 4.48, 4.739, 4.791, 5.213, 5.75, 5.79, 5.656, 5.464, 5.092, 3.135, 2.754, 3.14, 5.213, 5.831, 5.736, 5.333, 4.797, 6.704, 6.493, 5.746, 6.31, 6.53, 6.63, 6.071, 5.706, 4.951, 4.392, 4.619, 5.029, 5.302, 5.151, 3.474, 3.285, 3.232, 3.4, 3.503, 3.649, 3.828, 4.235, 4.781, 5.096, 5.262, 5.411, 5.439, 5.08, 4.719, 4.519)

length(ozone)
print(ozone)
par(mfrow = c(1, 2))
boxplot(ozone,main="boxplot for ozone sample")
abline(h = min(ozone), col = "Blue")
abline(h = max(ozone), col = "Yellow")
abline(h = median(ozone), col = "Green")
abline(h = quantile(ozone, c(0.25, 0.75)), col = "Red")
fivenum(ozone)

Boxplot:

output:

minimum=1.828

Q1=4.030

Q2=4.791

Q3=5.425

maximum=6.704

outlier seen from boxplot

Solution-2:

Rcode:

mean(ozone)
sd(ozone)

Output:

mean=4.716559

standard deviation=1.073852

Solution-3:

qqnorm(ozone)
qqline(ozone)

From qqplot ozone sample follows normal distribution

4. Are the assumptions for an hypothesis test of the mean reasonably satisfied?

Yes met since sample follows normal distribution

sample is random sample

and independent


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