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Katena32 [7]
2 years ago
6

The ages​ (years) of three government officials when they died in office were 56​, 46​, and 60. complete parts​ (a) through​ (d)

.a. assuming that 2 of the ages are randomly selected with​ replacement, list the different possible samples.
Mathematics
1 answer:
natita [175]2 years ago
6 0
There are 9 possible samples for selection of two with replacement. 

They are: (56, 56) (56, 46) (56, 60) (46, 46) (46, 56) (46, 60) (60, 60) (60, 46) (60, 56)

Remember that when making solution sets with replacement you have the possibility of choosing the same number (or data point) twice. So in this case, you will have the number of possible answers squared for total possible answers. 
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Suppose a given control unit fails, on average, every 12,000 hours. It takes an average of 900 hours to repair and reboot this u
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Answer:

The repair/reboot time needs to be reduced by 600 hours so as to increase availability time by 5%.

Step-by-step explanation:

To increase availability by 5%, the amount of hours of availability is multiplied 105%.

therefore 1.05 × 12000 = 12600 hours

Therefore the repair/reboot time needs to be reduced by 600 hours.

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2 years ago
A shoe manufacturer compared material A and material B for the soles of shoes. Twelve volunteers each got two shoes. The left wa
sveta [45]

Answer:

a) Are dependent since we are mesuring at the same individuals but on different times and with a different method

b) If we see the qq plot we don't have any significant deviation for the values and we don't have any heavy tail so we can conclude that we can approximate the differences with the normal distribution.

c) p_v =P(t_{(12)}>0.969) =0.353

So the p value is higher than the significance level given, so then we can conclude that we FAIL to reject the null hypothesis. So we can conclude that the mean differences is NOT significantly different from 0 .

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.  

The Q-Q plot, or quantile-quantile plot, "is a graphical tool to help us assess if a set of data plausibly came from some theoretical distribution such as a Normal or exponential".

Let put some notation  

x=value for A , y = value for B

A: 379, 378, 328, 372, 325, 304, 356, 309, 354, 318, 355, 392

B: 372, 376, 328, 368, 283, 252, 369, 321, 379, 303, 328, 411

(a) Are the two samples paired or independent? Explain your answer.

Are dependent since we are mesuring at the same individuals but on different times and with a different method

(b) Make a normal QQ plot of the differences within each pair. Is it reasonable to assume a normal population of differences?

The first step is calculate the difference d_i=A_i-B_i and we obtain this:

d: 7,2,0,4,42,52,-13,-12,-25,15,27,-19

In order to do the qqplot we can use the following R code:

d<-c(7,2,0,4,42,52,-13,-12,-25,15,27,-19)

qqnorm(d)

And the graph obtained is attached.

If we see the qq plot we don't have any significant deviation for the values and we don't have any heavy tail so we can conclude that we can approximate the differences with the normal distribution.

(c) Choose a test appropriate for the hypotheses above and justify your choice based on your answers to parts (a) and (b). Perform the test by computing a p-value, make a test decision, and state your conclusion in the context of the problem

The system of hypothesis for this case are:

Null hypothesis: \mu_A- \mu_B = 0

Alternative hypothesis: \mu_A -\mu_B \neq 0

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{80}{12}=6.67

The third step would be calculate the standard deviation for the differences, and we got:

s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =23.849

The 4 step is calculate the statistic given by :

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{6.67 -0}{\frac{23.849}{\sqrt{12}}}=0.969

The next step is calculate the degrees of freedom given by:

df=n-1=12-1=11

Now we can calculate the p value, since we have a two tailed test the p value is given by:

p_v =P(t_{(12)}>0.969) =0.353

So the p value is higher than the significance level given, so then we can conclude that we FAIL to reject the null hypothesis. So we can conclude that the mean differences is NOT significantly different from 0 .

4 0
2 years ago
Suppose a baseball team has 14 players on the roster who are not members of the pitching staff. Of those 14 players, assume that
Nadusha1986 [10]

Answer: Our required probability would be 0.70.

Step-by-step explanation:

Since we have given that

Number of players = 14

Number of players have recently taken a performance enhancing drug = 3

Number of players have not recently taken a performance enhancing drug = 14-3=11

Number of members chosen randomly = 5

We need to find the probability that at least one of the tested players is found to have taken a performance enhancing drug.

P(Atleast 1) = 1 - P(none is found to have taken a performance enhancing drug)

So, P(X≥1)=1-P(X=0)

P(X\geq 1)=1-^5C_0(\dfrac{11}{14})^5\\\\P(X\geq 1)=1-(0.786)^5\\\\P(X\geq 1)=0.70

Hence, our required probability would be 0.70.

7 0
2 years ago
Find the mean, median and mode of the weights of the people shown. 105kg 53kg 76kg 91kg 120kg 61kg 55kg 98kg 61kg
Yuliya22 [10]
First, you need to put them in order
53kg, 55kg, 61kg, 61kg, 76kg, 91kg, 98kg, 105kg, 120kg

For mean, you add them all up and divide by the amount of numbers (9)
720/9 = 80

For median, you find the middle number (76kg)

For mode, you find the number that appears the most (61kg)

Mean: 80
Median: 76
Mode: 61
5 0
2 years ago
Read 2 more answers
An observational study found that the amount of sleep an employee gets each night is associated with job performance. The correl
Elina [12.6K]

Answer:

The correct option is D.) Causation cannot be determined from an observational study.

Step-by-step explanation:

The conclusion is not correct because

D.) Causation cannot be determined from an observational study.

Causation determined from an observational study is speculative and cannot be confirmed without data from a real experiment.

6 0
2 years ago
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