Answer: Eric still needs of a deck.
Step-by-step explanation:
You know that:
- Eric need 1/2 deck of playing cards.
- Eric has 2/7 of a deck.
Therefore, to find the fraction of a deck that Eric still needs, you must subtract 1/2 and 2/7:
- Find the least common multiple of the denominators:
- Now, you can make the subtraction:
Therefore, he still needs of a deck.
Answer:
b. H0: μ < 530. Reject H0 if tcalc > -1.753
Step-by-step explanation:
1) Data given and notation
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)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is less than 530 (left tailed tes), the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, 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 if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
Critical value
Since we are conducting a left tailed test we need to first find the degrees of freedom for the statistic given by:
Now we need to look on the t distribution with 15 degrees of freedom that accumulates 0.05 of the area on the left area. And on this case the critical value would be .
And we can use the following excel code to find it: "=T.INV(0.05,15)"
So then the correct rejection zone for H0 would be: Reject H0 if
P-value
Since is a one side left tailed test the p value would be:
Conclusion
If we compare the p value and the significance level for example we see that
so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean is significantly lower than 530 at 5% of significance.
Answer:
Student Hours worked
April.
Debbie.
Richard.
Step-by-step explanation:
Some data's were missing so we have attached the complete information in the attachment.
Given:
Number of Hours Carl worked on Math project =
can be Rewritten as
Number of Hours Carl worked on Math project =
Number of Hours Sonia worked on Math project =
can be rewritten as
Number of Hours Sonia worked on Math project =
Number of Hours Tony worked on Math project =
can be rewritten as
Number of Hours Tony worked on Math project =
Now Given:
April worked times as long on her math project as did Carl.
can be Rewritten as
Number of Hours April worked on math project =
Number of Hours Carl worked on Math project
Number of Hours April worked on math project =
Also Given:
Debbie worked times as long as Sonia.
can be Rewritten as
.
Number of Hours Debbie worked on math project =
Number of Hours Sonia worked on Math project
Number of Hours Debbie worked on math project =
Also Given:
Richard worked times as long as tony.
can be Rewritten as
Number of Hours Richard worked on math project =
Number of Hours Tony worked on Math project
Number of Hours Debbie worked on math project =
Hence We will match each student with number of hours she worked.
Student Hours worked
April.
Debbie.
Richard.
