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Soloha48 [4]
4 years ago
11

Single-sample t test and paid days off: The number of paid days off (e.g., vacation, sick leave) taken by eight employees at a s

mall local business is compared to the national average. You are hired by the new business owner to help her determine what to expect for paid days off. In general, she wants to set some standard for her employees and for herself. Let's assume your search on the Internet for data on paid days off leaves you with the impression that the national average is 15 days. The data for the eight local employees during the last fiscal year are: 10, 11, 8, 14, 13, 12, 12, and 27 days.
A. Write hypothesis for your research.
B. Which type of test would be appropriate to analyze these data?
C. Before doing any computations, do you have any concerns about this research? Are there any questions you might like to ask about the data you have been given?
D. Calculate the appropriate t statistic.
Mathematics
1 answer:
SVEN [57.7K]4 years ago
6 0

Answer:

A. µ is the population average number of paid days off taken by employees that the company.

Null Hypothesis: µ=15 days

Alternate Hypothesis: µ≠15 days

B. one-sample t-test

C. We are unsure if it meets the conditions for a one-sample t-test

D. t=-0.79178; df=7

Step-by-step explanation:

For the null and alternate hypotheses, you want to first define what µ represents. Next, you state them, the null hypothesis being equal to the previously known average, and the alternate hypothesis being greater than, less than, or not equal to the previous average, selecting one depends on the context in the problem.

A one-mean t-test would be used as we are looking to compare the mean of a data set, as well as the fact that we do not know the population standard deviation.

When conducting these tests, we need to ensure that three conditions are being met. The first is random, which means that the data set is randomly gathered, or not, the data here does not seem to be random, which may be concerning. Next is independence, this is done when we survey less than 10% of the overall population, in this case it is a small company, so we do not know if it is less than 10% of the population. Last is normality, the data set is not sufficiently large (greater than 30 people surveyed) so we cannot use the central limit theorem to justify that the data is normal. We can use a normality plot, but when the data is placed on a normality plot most of the data appears to be linear, but the 27 day data point does not seem to be normal, so we cannot fully ensure that it is normal. Based on the data not following these conditions, we have concerns about proceeding with the test, we will therefore have to proceed with caution.

For the last part, use that T-test function on a calculator with statistics functions. Remember to include the degrees of freedom in the answer. (The degrees of freedom is one less than the sample size).

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Answer:

$22.34

Step-by-step explanation:

24 pennies is .24 of a dollar

62 nickels will be: (62 x 5) / 100 = 3.10

55 dimes is 5.5

16 quarters will be 4.0

19 fifty-cents will be 9.5

Add them up to get $22.34 if I’m right.

Hope this helped.

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Step-by-step explanation:

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The American Water Works Association reports that the per capita water use in a single-family home is 63 gallons per day. Legacy
Debora [2.8K]

Answer:

t=\frac{60-63}{\frac{8.9}{\sqrt{28}}}=-1.784    

df=n-1=28-1=27  

p_v =P(t_{(27)}  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is less than 63 gallons per day at 1% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=60 represent the sample mean

s=8.9 represent the sample standard deviation

n=28 sample size  

\mu_o =68 represent the value that we want to test

\alpha=0.01 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v 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 true mean is less than 63, the system of hypothesis would be:  

Null hypothesis:\mu \geq 63  

Alternative hypothesis:\mu < 63  

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:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (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:  

t=\frac{60-63}{\frac{8.9}{\sqrt{28}}}=-1.784    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=28-1=27  

Since is a one side lower test the p value would be:  

p_v =P(t_{(27)}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is less than 63 gallons per day at 1% of signficance.  

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Elza [17]
It is easier to compare a positive number and a negative number because it’s just like regular adding
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3 years ago
Can someone help me please .
Ymorist [56]

Answer:

D

Step-by-step explanation:

The angle CDG equals 90 degrees and the angle GDE also equals 90 degrees. Supplementary angles are angles that add up to be 180 degrees.

CDG + GDE = 180 degrees

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3 years ago
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