A student examines 27 geological samples for nitrate concentration. The mean nitrate concentration for the sample data is 0.819 cc/cubic meter with a standard deviation of 0.0881. Determine the 90% confidence interval for the population mean lead concentration. Assume the population is approximately normal.Step 1 of 2: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places. Construct the 90% confidence interval. Round your answer to three decimal places

Accepted Solution

Answer: 90% confidence interval: (0.790, 0.848)Critical t- value for 90% confidence = Β 1.706Step-by-step explanation:As we consider the given description, we hvaen= 27[tex]\overline{x}=0.819[/tex]s=0.0881Since population standard deviation is unknown.so we use t-critical valueUsing t-value table , the critical t- value will be:-[tex]t_{n-1,\ \alpha/2}=t_{26,\ 0.05}= 1.706[/tex]Confidence interval : [tex]\overline{x}\pm t_{n-1,\alpha/2}\dfrac{s}{\sqrt{n}}[/tex]i.e. [tex]0.819\pm ( 1.706)\dfrac{0.0881}{\sqrt{27}}[/tex][tex]\approx0.819\pm 0.029=(0.819-0.029,\ 0.819+0.029)\\\\=(0.790,\ 0.848)[/tex]Hence, 90% confidence interval: (0.790, 0.848)