Answer:
a. [6.6350,7.3950]
b. ME=0.5150
Step-by-step explanation:
a. Given that n=40, [tex]\bar x=6.88[/tex] and that:[tex]z_{\alpha/2}=z_{0.05}=1.645[/tex]
The required 90% confidence interval can be calculated as:
[tex]\bar x\pm(margin \ of \ error)\\\\\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\\\6.88\pm(1.645\times \frac{1.98}{\sqrt{40}})\\\\=[6.3650,7.3950][/tex]
Hence, the 90% confidence interval for the population mean cash value of this crop is [6.6350,7.3950]
b. The margin of error at 90% confidence interval is calculated as:
[tex]ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\\\=(1.645\times \frac{1.98}{\sqrt{40}})\\\\=0.5150[/tex]
Hence, the margin of error is 0.5150
