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Sagot :
Answer:
The 90% confidence interval for the mean difference in pulse rates for the population of study participants is between -2.9 beats per minute and 0.5 beats per minute.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 10 - 1 = 9
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.9}{2} = 0.95[/tex]. So we have T = 1.8331
The margin of error is:
[tex]M = T\frac{s}{\sqrt{n}} = 1.8331\frac{2.97}{\sqrt{10}} = 1.7[/tex]
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is -1.2 - 1.7 = -2.9 beats per minute
The upper end of the interval is the sample mean added to M. So it is -1.2 + 1.7 = 0.5 beats per minute.
The 90% confidence interval for the mean difference in pulse rates for the population of study participants is between -2.9 beats per minute and 0.5 beats per minute.
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