peter runs laps on the school track every afternoon. he wants to run no more than 45 minutes a day. he can consistently run each
1 answer:
Answer:
see the explanation
Step-by-step explanation:
Let
x ----> the number of laps
we know that
The number of laps multiplied by 1.5 minutes each must be at most 45 minutes.
The term "at most" means "less than or equal to"
The inequality that represent this situation is
solve for x
Divide by 1.5 both sides
The maximum number of laps is 30
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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.
