Daniel is on an airplane flying to a destination 825 miles away. After one hour, the pilot announces that the plane is 495 miles
2 answers:
Total distance of the destination = 825 miles
After one hour distance covered = miles
After one hour again the distance covered = miles
Total distance covered till this time = miles
Now, the equation that represents the remaining distance, in miles, is option B:
d = 825 – 330t (330 multiplied by number of hours as the speed is constant)
or check this by putting t=2
d = 825 - 330(2)
d = 825 - 660
d = 165 miles left.
Hence, option B is correct.
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Answer:
1) H0: There is independence between the marital status and the diagnostic of alcoholic
H1: There is association between the marital status and the diagnostic of alcoholic
2) The statistic to check the hypothesis is given by:
3)
4)
And we can calculate the p value given by:
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(19.72,2,TRUE)"
Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.
Step-by-step explanation:
A chi-square goodness of fit test "determines if a sample data matches a population".
A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".
Assume the following dataset:
Diag. Alcoholic Undiagnosed Alcoholic Not alcoholic Total
Married 21 37 58 116
Not Married 59 63 42 164
Total 80 100 100 280
Part 1
We need to conduct a chi square test in order to check the following hypothesis:
H0: There is independence between the marital status and the diagnostic of alcoholic
H1: There is association between the marital status and the diagnostic of alcoholic
The level os significance assumed for this case is
Part 2
The statistic to check the hypothesis is given by:
Part 3
The table given represent the observed values, we just need to calculate the expected values with the following formula
And the calculations are given by:
And the expected values are given by:
Diag. Alcoholic Undiagnosed Alcoholic Not alcoholic Total
Married 33.143 41.429 41.429 116
Not Married 46.857 58.571 58.571 164
Total 80 100 100 280
And now we can calculate the statistic:
Part 4
Now we can calculate the degrees of freedom for the statistic given by:
And we can calculate the p value given by:
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(19.72,2,TRUE)"
Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.