The sample data below shows the number of hours spent by five students over the weekend to prepare for Monday's Business Statist
ics exam. 3 12 2 3 5(See the Excel Data File.) The 75 th percentile of the data is the closest to ________.g
1 answer:
Answer:
8.5 hrs
Step-by-step explanation:
-A 75th percentile mathematically means that 75% of the time data points are below that value and 25% of the time are above that value.
-We plot the given our data {3 12 2 3 5} to determine the 75th percentile
#We can use to plot our boxplot.
-From the plot, we find our 75th percentile to be 8.5 hrs
Hence, 75% of the time the number of hours spent to prepare was less than 8.5hrs.
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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.