Answer :
Answer:
(a) The 80% confidence interval for the population mean nitrate concentration is (0.648, 0.692).
(b) The critical value of t that should be used in constructing the 80% confidence interval is 1.345.
Step-by-step explanation:
Let X = nitrate concentration.
The sample mean nitrate concentration is, [tex]\bar x=0.670[/tex] cc/cubic meter.
The sample standard deviation of the nitrate concentration is, [tex]s=0.0616[/tex].
It assumed that the population is approximately normal.
And since the population standard deviation is not known, we will use a t-interval.
The (1 - α)% confidence interval for population mean (μ) is:
[tex]CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}[/tex]
(a)
The critical value of t for α = 0.20 and degrees of freedom, (n - 1) = 14 is:
[tex]t_{\alpha/2, (n-1)}=t_{0.20/2, (15-1)}=t_{0.10, 14}=1.345[/tex]
*Use a t-table for the critical value.
Compute the 80% confidence interval for the population mean nitrate concentration as follows:
[tex]CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}[/tex]
[tex]=0.670\pm 1.345\times \frac{0.0616}{\sqrt{15}}[/tex]
[tex]=0.670\pm 0.022\\=(0.648, 0.692)\\[/tex]
Thus, the 80% confidence interval for the population mean nitrate concentration is (0.648, 0.692).
(b)
The critical value of t for confidence level (1 - α)% and (n - 1) degrees of freedom is:
[tex]t_{\alpha/2, (n-1)}[/tex]
The value of is:
α = 0.20
And the degrees of freedom is,
(n - 1) = 15 - 1 = 14
Compute the critical value of t for confidence level 80% and 14 degrees of freedom as follows:
[tex]t_{\alpha/2, (n-1)}=t_{0.20/2, (15-1)}[/tex]
[tex]=t_{0.10, 14}\\=1.345[/tex]
*Use a t-table for the critical value.
Thus, the critical value of t that should be used in constructing the 80% confidence interval is 1.345.
