Question

Grandma​ Gertrude's Chocolates, a family owned​ business, has an opportunity to supply its product for distribution through a large coffee house chain.​ However, the coffee house chain has certain specifications regarding cacao content as it wishes to advertise the health benefits​ (antioxidants) of the chocolate products it sells. In order to determine the mean percentage of cacao in its dark chocolate​ products, quality inspectors sample 36 pieces. They find a sample mean of​ 60% with a standard deviation of​ 8%. What is the correct value of t to construct a​ 90% confidence interval for the true mean percentage of​ cacao? (Round to the nearest thousandth.)
Select Answer:
A. (0.577, 0.623)
B. (0.539, 0.561)
C. (0.527, 0.573)
D. (0.589, 0.611)

Answer 1

Answer with explanation:

Given : Sample size : n = 36

Significance level for 90% confidence : $\alpha: 1-0.9=0.1$

Sample mean : $\overline{x}=0.60$

Standard deviation : $\sigma=0.08$

By using standard normal table for t-values,

Critical t-value : $t_{(n-1, \alpha/2)}=t_{(35,\ 0.05)}=1.690$

Thus, the correct value of t to construct a​ 90% confidence interval for the true mean percentage of​ cacao : $t_{(35,\ 0.05)}=1.690$

Confidence interval for population mean :

$\overline{x}\pm t_{(n-1, \alpha/2)}\dfrac{\sigma}{\sqrt{n}}\\\\=0.60\pm(1.69)\dfrac{0.08}{\sqrt{36}}\\\\=0.60\pm0.0225333333333\\\\\approx0.60\pm0.023\\\\=(0.60-0.023,\ 0.60+0.023)=(0.577,\ 0.623)$

Hence, A is the correct answer .