Given : Sample size : n = 36
Significance level for 90% confidence : [tex]\alpha: 1-0.9=0.1[/tex]
Sample mean : [tex]\overline{x}=0.60[/tex]
Standard deviation : [tex]\sigma=0.08[/tex]
By using standard normal table for t-values,
Critical t-value : [tex]t_{(n-1, \alpha/2)}=t_{(35,\ 0.05)}=1.690[/tex]
Thus, the correct value of t to construct a 90% confidence interval for the true mean percentage of cacao : [tex]t_{(35,\ 0.05)}=1.690[/tex]
Confidence interval for population mean :
[tex]\overline{x}\pm t_{(n-1, \alpha/2)}\dfrac{\sigma}{\sqrt{n}}\\\\=0.60\pm(1.69)\dfrac{0.08}{\sqrt{36}}\\\\=0.60\pm0.0225333333333\\\\\approx0.60\pm0.023\\\\=(0.60-0.023,\ 0.60+0.023)=(0.577,\ 0.623)[/tex]
Hence, A is the correct answer .
