Respuesta :
Answer:
a) The 98% confidence interval for the unknown population mean dissolved oxygen content is between 8.44 mg/L and 9.84 mg/L.
b) This interval means that we are 98% sure that the true population mean dissolved oxygen content is between 8.44 mg/L and 9.84 mg/L.
c) A sample size of 87 is required to get a 98% confidence interval with error margin 0.5
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.98}{2} = 0.01[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.01 = 0.99[/tex], so [tex]z = 2.327[/tex]
Now, find M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
[tex]M = 2.327*\frac{2}{\sqrt{44}} = 0.7[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 9.14 - 0.7 = 8.44 mg/L.
The upper end of the interval is the sample mean added to M. So it is 9.14 + 0.7 = 9.84 mg/L.
The 98% confidence interval for the unknown population mean dissolved oxygen content is between 8.44 mg/L and 9.84 mg/L.
b) This interval means that we are 98% sure that the true population mean dissolved oxygen content is between 8.44 mg/L and 9.84 mg/L.
(c) What sample size n is required to get a 98% confidence interval with error margin 0.5?
This is n for which M is 0.5. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.5 = 2.327*\frac{2}{\sqrt{n}}[/tex]
[tex]0.5\sqrt{n} = 2.327*2[/tex]
[tex]0.5\sqrt{n} = 4.654[/tex]
[tex]\sqrt{n} = \frac{4.654}{0.5}[/tex]
[tex]\sqrt{n} = 9.308[/tex]
[tex](\sqrt{n})^{2} = (9.308)^{2}[/tex]
[tex]n = 86.6[/tex]
Rouding up
A sample size of 87 is required to get a 98% confidence interval with error margin 0.5
