In an experiment to determine the effect of ambient temperature on the emissons of oxides of nitrogen ( NOx ) of diesel trucks, 10 trucks were run at temperatures of 40°F and 80°F . The emissions, in parts per billion, are presented in the following table. Truck 40°F 80°F 1 926.5 896.7 2 851.1 857.0 3 975.5 952.1 4 1009.3 884.8 5 871.8 840.7 6 949.2 885.1 7 1006.3 885.5 8 836.5 777.8 9 837.8 850.2 10 958.9 882.1 Send data to Excel Let μ1 represent the mean emission at 40°F and =μd−μ1μ2 . Can you conclude that the mean emission differs between the two temperatures? Use the =α0.05 level of significance and the TI-84 Plus calculator to answer the following. p value ? do we reject? is there enough evidence :?
Inverse Normal Distribution
The method used for finding the corresponding z-critical value in a normal distribution using the known probability is said to be an inverse normal distribution. The inverse normal distribution is a continuous probability distribution with a family of two parameters.
Mean, Median, Mode
It is a descriptive summary of a data set. It can be defined by using some of the measures. The central tendencies do not provide information regarding individual data from the dataset. However, they give a summary of the data set. The central tendency or measure of central tendency is a central or typical value for a probability distribution.
Z-Scores
A z-score is a unit of measurement used in statistics to describe the position of a raw score in terms of its distance from the mean, measured with reference to standard deviation from the mean. Z-scores are useful in statistics because they allow comparison between two scores that belong to different normal distributions.
In an experiment to determine the effect of ambient temperature on the emissons of oxides of nitrogen (
) of diesel trucks,
trucks were run at temperatures of
and
. The emissions, in parts per billion, are presented in the following table.
| Truck |
40°F
|
80°F
|
|
1
|
926.5
|
896.7
|
|
2
|
851.1
|
857.0
|
|
3
|
975.5
|
952.1
|
|
4
|
1009.3
|
884.8
|
|
5
|
871.8
|
840.7
|
|
6
|
949.2
|
885.1
|
|
7
|
1006.3
|
885.5
|
|
8
|
836.5
|
777.8
|
|
9
|
837.8
|
850.2
|
|
10
|
958.9
|
882.1
|
Let
represent the mean emission at
and
.
Can you conclude that the mean emission differs between the two temperatures? Use the
level of significance and the TI-84 Plus calculator to answer the following.
p value ?
do we reject?
is there enough evidence :?
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