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lianna [129]
3 years ago
5

A study looked at n=238 adolescents, all free of severe illness.23 Subjects wore a wrist actigraph, which allowed the researcher

s to estimate sleep patterns. Those subjects classified as having low sleep efficiency had an average systolic blood pressure that was 5.8 millimeters of mercury (mm Hg) higher than that of other adolescents. The standard deviation of this difference is 1.4 mm Hg. Based on these results, test whether this difference is significant at the 0.01 level.
Mathematics
1 answer:
Vesnalui [34]3 years ago
8 0

Answer:

Step-by-step explanation:

Since n < 30, we will find the t-score and compare that to the t-score of a significance level of 1%.    

Since they are asking if the difference is significant, we will have a two tailed test, with degree of freedom being 22, so our critical values are

t < -2.704 and t > 2.704

Our t-value for this situation is

t = ([µ + 5.8] - µ)/(1.4/√23)

It's µ + 5.8 because the problem told us that their levels are 5.8 mm higher than the average, so it's the average, plus 5.8

Simplify the equation...

t = 5.8/(1.4/√23)

t = 19.868

19.868 > 2.704, the evidence supports that there is significant difference between the sample and the population

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3 years ago
Can someone show me how to solve this
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Answer:

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The above mentioned are false statements

Step-by-step explanation:

f(1)=13 : false because "13" is out of range -40≤f(x)≤-10

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7 0
3 years ago
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Answer:

132

Step-by-step explanation:

<em>Theorem:</em>

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8 0
3 years ago
Current research indicates that the distribution of the life expectancies of a certain protozoan is normal with a mean of 48 day
irina1246 [14]

Answer:

P ( x_bar ≥ 51 ) = 0.0432

Step-by-step explanation:

Solution:-

- The random variable "X" denotes:

                  X : life expectancies of a certain protozoan

- The variable "X" follows normal distribution.

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- A sample of n = 36 days was taken.

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                  x ~ Norm ( 48 , ( 10.5 / √n)^2 )

- The sample standard deviation s = 10.5 / √n = 10.5 / √36

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- We are to investigate the the probability of sample mean x_bar ≥ 51 days:

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- Standardize the results, evaluate Z-score:

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- Use the standardized normal table and evaluate:

                 P ( Z ≥ 1.7142 ) = 0.0432                  

Hence,      P ( x_bar ≥ 51 ) = 0.0432

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3 years ago
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