Not the best at this! Can someone please help me ?!
1 answer:
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Answer:
If we compare the p value and the significance level assumed we see that
so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude the the true mean for method 1 is lower than the mean for the method 2 at 5% of significance
Step-by-step explanation:
Data given and notation
We can calculate the sample mean and deviation with these formulas:
represent the mean for the sample mean for 1
represent the mean for the sample mean for 2
represent the sample standard deviation for the sample 1
represent the sample standard deviation for the sample 2
sample size selected 1
sample size selected 2
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)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the average time taken when training under method 1 is less than the average time for Method 2, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:
(1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".
Calculate the statistic
We can replace in formula (1) the info given like this:
P-value
The first step is calculate the degrees of freedom, on this case:
Since is a one sided test the p value would be:
Conclusion
If we compare the p value and the significance level assumed we see that
so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude the the true mean for method 1 is lower than the mean for the method 2 at 5% of significance
A) Since a is last in line, we can disregard a, and concentrate on the remaining letters.
Let's start by drawing out a representation:
_ _ _ _ a
Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways
b) Since a is before d, we need to account for all of the possible cases.
Case 1: a d _ _ _
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d
Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.
a b c _ _
We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:
_ a b c _
_ _ a b c
So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.
In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).
Let's start by drawing out a representation:
_ _ _ _ a
Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways
b) Since a is before d, we need to account for all of the possible cases.
Case 1: a d _ _ _
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d
Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.
a b c _ _
We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:
_ a b c _
_ _ a b c
So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.
In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).