What is 2(3/5×15) ????????
2 answers:
You might be interested in
Answer:
The probability of using one or the other is 36%
Step-by-step explanation:
For solving this problem it is easy if we see it in a ven diagram, for this first we are going to name the initial conditions with some variables:
Probability of passing Professor Jones math class = 64% =0,64
P(J) = 0.64
Probabiliry of passing Professor Smith's physics class = 32% =0.32
P(S) = 0.32
Probability of passing both is = 30% = 0.30
P(JnS) = 0.30 (Is is an intersection so it is in the middle of the ven diagram
We need to know which is the probability of pasing one or the other for this we need to take out the probability of passing both for this we have to add the probability of passing Professor Jones math class with the probabiliry of passing Professor Smith's physics class and substract the probability of passing both for each one:
P(JuS) = (P(J) - P(JnS)) + (P(S) - P(JnS)) = (0.64 - 0.30) + (0.32 - 0.30) = 0.34 + 0.02 = 0.36 = 36%
If you check the ven diagram you can see that if we add all what is in red we will have the probability of passing Professor Jones math class and if we add all what is in blue we wiill have the probability of passing Professor Smith's physics class, and if we add just what is in each corner we will get the same value that is the probabilty of passsing one or the other.
Answer:
So the p value obtained was a very low value and using the significance level given we have
so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that regularly watch the TV station’s news program is significantly less than 0.38 .
Step-by-step explanation:
1) Data given and notation
n=830 represent the random sample taken
X=282 represent the people that regularly watch the TV station’s news program
estimated proportion of people that regularly watch the TV station’s news program
is the value that we want to test
represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the proportion is less than 0.38:
Null hypothesis:
Alternative hypothesis:
When we conduct a proportion test we need to use the z statisitc, and the is given by:
(1)
The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value
.
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level provided . The next step would be calculate the p value for this test.
Since is a left tailed test the p value would be:
So the p value obtained was a very low value and using the significance level given we have
so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that regularly watch the TV station’s news program is significantly less than 0.38 .