Find the mean median mode of 12,25,23,17,23
2 answers:
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Answer:
Step-by-step explanation:
Data given and notation
represent the sample mean
represent the sample standard deviation for the sample
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 weight is less than 4 ounces, 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:
Note that 6% converted to a decimal number is 6/100=0.06. Also note that 6% of a certain quantity x is 0.06x.
Here is how much the worker earned each year:
In the year 1985 the worker earned <span>$10,500.
</span>In the year 1986 the worker earned $10,500 + 0.06($10,500). Factorizing $10,500, we can write this sum as:
$10,500(1+0.06).
In the year 1987 the worker earned
$10,500(1+0.06) + 0.06[$10,500(1+0.06)].
Now we can factorize $10,500(1+0.06) and write the earnings as:
$10,500(1+0.06) [1+0.06]=
.
Similarly we can check that in the year 1987 the worker earned
, which makes the pattern clear.
We can count that from the year 1985 to 1987 we had 2+1 salaries, so from 1985 to 2010 there are 2010-1985+1=26 salaries. This means that the total paid salaries are:
.
Factorizing, we have
![=10,500[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]=10,500\cdot[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]](https://tex.z-dn.net/?f=%3D10%2C500%5B1%2B1.06%2B%281.06%29%5E2%2B%281.06%29%5E3%2B...%2B%281.06%29%5E%7B26%7D%5D%3D10%2C500%5Ccdot%5B1%2B1.06%2B%281.06%29%5E2%2B%281.06%29%5E3%2B...%2B%281.06%29%5E%7B26%7D%5D)
We recognize the sum as the geometric sum with first term 1 and common ratio 1.06, applying the formula
(where a is the first term and r is the common ratio) we have:
.
Finally, multiplying 10,500 by 59.17 we have 621.285 ($).
The answer we found is very close to D. The difference can be explained by the accuracy of the values used in calculation, most important, in calculating
.
Answer: D
Here is how much the worker earned each year:
In the year 1985 the worker earned <span>$10,500.
</span>In the year 1986 the worker earned $10,500 + 0.06($10,500). Factorizing $10,500, we can write this sum as:
$10,500(1+0.06).
In the year 1987 the worker earned
$10,500(1+0.06) + 0.06[$10,500(1+0.06)].
Now we can factorize $10,500(1+0.06) and write the earnings as:
$10,500(1+0.06) [1+0.06]=
Similarly we can check that in the year 1987 the worker earned
We can count that from the year 1985 to 1987 we had 2+1 salaries, so from 1985 to 2010 there are 2010-1985+1=26 salaries. This means that the total paid salaries are:
Factorizing, we have
We recognize the sum as the geometric sum with first term 1 and common ratio 1.06, applying the formula
Finally, multiplying 10,500 by 59.17 we have 621.285 ($).
The answer we found is very close to D. The difference can be explained by the accuracy of the values used in calculation, most important, in calculating
Answer: D