Answer:
Since the calculated values is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level of 2.5% and we can say that the true mean is lower than 36 years old
Step-by-step explanation:
Data given
represent the sample mean
represent the sample standard deviation
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
System of hypothesis
We need to conduct a hypothesis in order to check if the true mean is less than 36 years old, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
The statistic is given by:
(1)
And replacing we got:
Now we can calculate the critical value but first we need to find the degreed of freedom:
So we need to find a critical value in the t distribution with df =21 who accumulates 0.025 of the area in the left and we got:
Since the calculated values is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level of 2.5% and we can say that the true mean is lower than 36 years old
15x.25=3.75
15-3.75=11.25.
The answer is $11.25
2.1 x 82 =
172.2
I hope this answer was helpful! :)
Answer:
What do you mean by that?
Step-by-step explanation:
Answer:
25%
Step-by-step explanation:
This question is about conditional probability. Let's say that the probability of raining on Saturday is X=true and the probability of raining on Sunday is Y=true. There is a 15% it will rain on both Saturday and Sunday, to put into the equation it will be:
P(X= true ∩ Y = true) = P(X = true) * P(Y = true)= 0.15
There is a 60% chance of rain on Saturday, mean the equation is
P(X = true) = 0.6
The question is asking for the chance of rain on Sunday or P(Y = true). If we substitute the second equation to first, it will be:
P(X = true) * P(Y = true)= 0.15
0.6* P(Y = true)= 0.15
P(Y = true)= 0.15/0.6
P(Y = true)= 0.25 = 25%
