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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
There is a significant difference between the sample and the population.
What is standard deviation?
A standard deviation (or σ) indicates how dispersed the data is in relation to the mean.
What is t-distribution?
The t-distribution indicates the standardized distances of sample means to the population mean when the population standard deviation is not known, and the observations come from a normally distributed population.
What is t-score?
A t-score is the number of standard deviations away from the mean of the t-distribution.
How to find whether the difference is significant at the 0.01 level?
- Here n < 30.
- So, we will find the t-score and compare that to the t-score of a significance level of 1%.
Here the degree of freedom is 22 and so the critical values are:
t < -2.704 and t > 2.704
- In the problem, it is given that 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 t-value for this situation is
t = ([µ + 5.8] - µ)/(1.4/√23)
⇒ t = 5.8/(1.4/√23)
⇒t = 19.868
- Now, 19.868 > 2.704
So, we can say that there is significant difference between the sample and the population.
Find out more details about "t-distribution" here: https://brainly.com/question/16959076
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