Respuesta :
Answer:
Part (a): 194; part (b): 90 and part (c): larger than
Step-by-step explanation:
Part (a): Range is 109-41 = 68
Std. deviation = range/4 = 68/4 = 17
For n, Error (E) = z of 90%*std. dev/[tex]\sqrt{x}[/tex]n
So n = (z of 90%*std. dev/E)^square...(i)
Z value for 90% confidence = 1-alpha = 0.90
alpha = 0.1
We take alpha/2 for z value calculation, so 0.1/2 is 0.05. Look up the z table and you get 1.64.
Going back to equation (i),
n = (z of 90%*std. dev/E)^square = (1.64*17/2)^square = 194.32 = 194
N must be a whole number that is why we have approximated it to 194.
Part (b): Put given values in equation (i)
n = (z of 90%*std. dev/E)^square = (1.64*11.6/2)^sqaure = 90.478 =90.5 = 90
0.5 is rounded down if the number on the left is an even number and is rounded up when the number on the left is odd. Here, 90 is an even number so we rounded down.
Part (c): The result from part (a)_________the result from part (b).
Since, 194 is 'larger than' 90 hence, result from part (a) is larger than the result from part (b). Also, result form part (a) is likely to be better as larger sample size gives more reliable results.
Based on the range rule of thumb, the sample size is equal to 194.
Given the following data:
- Population = 155 adult males.
- Low rate = 41 bpm.
- High rate = 109 bpm.
- Sample mean = 2 bpm.
- Confidence interval = 90%.
How to calculate the sample size.
First of all, we would determine the range and then use the range rule of thumb to estimate the standard deviation:
Note: Range represents the minimum and maximum value of a data set.
Range = [tex]109-41[/tex]
Range = 68.
For the standard deviation:
[tex]\delta =\frac{Range}{4} \\\\\delta =\frac{68}{4}[/tex]
Standard deviation = 17.
At 90% confidence interval, we have;
[tex]0.90=1-\alpha \\\\\alpha =1-0.90\\\\\alpha =0.10\\\\\\Z=\frac{\alpha}{2} \\\\Z =\frac{0.10}{2} \\\\Z =0.05[/tex]
From the z-table, we have;
Z-score = 1.64
Error = 2.
[tex]Error =Z_o\times \frac{\delta}{\sqrt{n} } \\\\\sqrt{n} =Z_o\times \frac{\delta}{E}\\\\n=(Z_o\times \frac{\delta}{E})^2\\\\n=(1.64 \times \frac{17}{2})^2\\\\n=(13.94)^2[/tex]
Sample size, n = 194.32 ≈ 194.
Part b.
When standard deviation = 11.6 bpm;
[tex]Error =Z_o\times \frac{\delta}{\sqrt{n} } \\\\\sqrt{n} =Z_o\times \frac{\delta}{E}\\\\n=(Z_o\times \frac{\delta}{E})^2\\\\n=(1.64 \times \frac{11.6}{2})^2\\\\n=(9.512)^2[/tex]
Sample size, n = 90.48 ≈ 90.
Part c.
Based on the calculations, the result from part (a) is larger than the result from part (b). Also, the result form part (a) is likely to be better because larger sample size gives more reliable results.
Read more on z-scores here: https://brainly.com/question/4302527