Nour drove from the Dead Sea up to Amman, and her altitude changed at a constant rate. When she began driving, her altitude was
1 answer:
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Given the table below of the prices for the Lenovo zx-81 chip during the last 12 months
![\begin{tabular} {|c|c|c|c|} Month&Price per Chip&Month&Price per Chip\\[1ex] January&\$1.90&July&\$1.80\\ February&\$1.61&August&\$1.83\\ March&\$1.60&September&\$1.60\\ April&\$1.85&October&\$1.57\\ May&\$1.90&November&\$1.62\\ June&\$1.95&December&\$1.75 \end{tabular}](https://tex.z-dn.net/?f=%5Cbegin%7Btabular%7D%0A%7B%7Cc%7Cc%7Cc%7Cc%7C%7D%0AMonth%26Price%20per%20Chip%26Month%26Price%20per%20Chip%5C%5C%5B1ex%5D%0AJanuary%26%5C%241.90%26July%26%5C%241.80%5C%5C%0AFebruary%26%5C%241.61%26August%26%5C%241.83%5C%5C%0AMarch%26%5C%241.60%26September%26%5C%241.60%5C%5C%0AApril%26%5C%241.85%26October%26%5C%241.57%5C%5C%0AMay%26%5C%241.90%26November%26%5C%241.62%5C%5C%0AJune%26%5C%241.95%26December%26%5C%241.75%0A%5Cend%7Btabular%7D)
The forcast for a period
is given by the formular
where
is the actual value for the preceding period and 
is the forcast for the preceding period.
Part 1A:
Given <span>α = 0.1 and the initial forecast for october of $1.83, the actual value for october is $1.57.
Thus, the forecast for period 11 is given by:
Therefore, the foreast for period 11 is $1.80
Part 1B:
</span>Given <span>α = 0.1 and the forecast for november of $1.80, the actual value for november is $1.62
Thus, the forecast for period 12 is given by:
Therefore, the foreast for period 12 is $1.78</span>
Part 2A:
Given <span>α = 0.3 and the initial forecast for october of $1.76, the actual value for October is $1.57.
Thus, the forecast for period 11 is given by:
Therefore, the foreast for period 11 is $1.70
</span>
<span><span>Part 2B:
</span>Given <span>α = 0.3 and the forecast for November of $1.70, the actual value for november is $1.62
Thus, the forecast for period 12 is given by:
Therefore, the foreast for period 12 is $1.68
</span></span>
<span>Part 3A:
Given <span>α = 0.5 and the initial forecast for october of $1.72, the actual value for October is $1.57.
Thus, the forecast for period 11 is given by:
Therefore, the forecast for period 11 is $1.65
</span>
<span><span>Part 3B:
</span>Given <span>α = 0.5 and the forecast for November of $1.65, the actual value for November is $1.62
Thus, the forecast for period 12 is given by:
Therefore, the forecast for period 12 is $1.64
Part 4:
The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.
Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75
using </span></span></span><span>α = 0.3, we obtained that the forcasted values of october, november and december are: $1.83, $1.80, $1.78
Thus, the mean absolute deviation is given by:
Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.1 of October, November and December is given by: 0.157
</span><span><span>Part 5:
The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.
Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75
using </span><span>α = 0.3, we obtained that the forcasted values of october, november and december are: $1.76, $1.70, $1.68
Thus, the mean absolute deviation is given by:
Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.3 of October, November and December is given by: 0.107
</span></span>
<span><span>Part 6:
The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.
Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75
using </span><span>α = 0.5, we obtained that the forcasted values of october, november and december are: $1.72, $1.65, $1.64
Thus, the mean absolute deviation is given by:
Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.5 of October, November and December is given by: 0.097</span></span>
The forcast for a period
where
Part 1A:
Given <span>α = 0.1 and the initial forecast for october of $1.83, the actual value for october is $1.57.
Thus, the forecast for period 11 is given by:
Therefore, the foreast for period 11 is $1.80
Part 1B:
</span>Given <span>α = 0.1 and the forecast for november of $1.80, the actual value for november is $1.62
Thus, the forecast for period 12 is given by:
Therefore, the foreast for period 12 is $1.78</span>
Part 2A:
Given <span>α = 0.3 and the initial forecast for october of $1.76, the actual value for October is $1.57.
Thus, the forecast for period 11 is given by:
Therefore, the foreast for period 11 is $1.70
</span>
<span><span>Part 2B:
</span>Given <span>α = 0.3 and the forecast for November of $1.70, the actual value for november is $1.62
Thus, the forecast for period 12 is given by:
Therefore, the foreast for period 12 is $1.68
</span></span>
<span>Part 3A:
Given <span>α = 0.5 and the initial forecast for october of $1.72, the actual value for October is $1.57.
Thus, the forecast for period 11 is given by:
Therefore, the forecast for period 11 is $1.65
</span>
<span><span>Part 3B:
</span>Given <span>α = 0.5 and the forecast for November of $1.65, the actual value for November is $1.62
Thus, the forecast for period 12 is given by:
Therefore, the forecast for period 12 is $1.64
Part 4:
The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.
Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75
using </span></span></span><span>α = 0.3, we obtained that the forcasted values of october, november and december are: $1.83, $1.80, $1.78
Thus, the mean absolute deviation is given by:
Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.1 of October, November and December is given by: 0.157
</span><span><span>Part 5:
The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.
Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75
using </span><span>α = 0.3, we obtained that the forcasted values of october, november and december are: $1.76, $1.70, $1.68
Thus, the mean absolute deviation is given by:
Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.3 of October, November and December is given by: 0.107
</span></span>
<span><span>Part 6:
The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.
Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75
using </span><span>α = 0.5, we obtained that the forcasted values of october, november and december are: $1.72, $1.65, $1.64
Thus, the mean absolute deviation is given by:
Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.5 of October, November and December is given by: 0.097</span></span>
Answer:
There is no significant difference between the two averages at 5% level
Step-by-step explanation:
Given that a a college student is interested in testing whether business majors or liberal arts majors are better at trivia.
The student gives a trivia quiz to a random sample of 30 business school majors and finds the sample’s average test score is 86. He gives the same quiz to 30 randomly selected liberal arts majors and finds the sample’s average quiz score is 89
Thus he has done a hypothesis testing for comparison of two means of different subjects. n =30
Since which is better is not claimed we use two tailed test here
We find that p value our alpha
Since p >alpha, we find that there is no significant difference between the averages of these two groups and null hypothesis is accepted