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​ % cacao in its dark chocolate​ products, quality inspectors sample 36 pieces. They find a sample mean of​ 55% with a standard deviation of​ 4%. The correct value of tSuperscript times to construct a​ 90% confidence interval for the true mean​ % cacao is​ _______.a.53.87% to 56.13%.
b.53.64% to 56.36%.
c.54.33% to 55.67%.
d.53.33% to 56.67%.
e.51% to 59%.

Answer 1

Answer:

90% confidence interval for the true mean % cacao is [53.87% , 56.13%].

Step-by-step explanation:

We are given that in order to determine the mean​ % cacao in its dark chocolate​ products, quality inspectors sample 36 pieces.

They find a sample mean of​ 55% with a standard deviation of​ 4%.

Firstly, the pivotal quantity for 90% confidence interval for the true mean is given by;

                            P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~ $t_n_-_1$

where, $\bar X$ = sample mean % cacao = 55%

             s = sample standard deviation = 4%

             n = sample of pieces = 36

             $\mu$ = true mean % cacao

Here for constructing 90% confidence interval we have used One-sample t test statistics as we know don't about population standard deviation.

So, 90% confidence interval for the true mean, $\mu$ is ;

P(-1.6895 < $t_3_6$ < 1.6895) = 0.90  {As the critical value of t at 35 degree of

                                           freedom are -1.6895 & 1.6895 with P = 5%}  

P(-1.6895 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.6895) = 0.90

P( $-1.6895 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.6895 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.6895 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.6895 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.6895 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.6895 \times {\frac{s}{\sqrt{n} } }$ ]

                    = [ $0.55-1.6895 \times {\frac{0.04}{\sqrt{36} } }$ , $0.55+1.6895 \times {\frac{0.04}{\sqrt{36} } }$ ]

                    = [0.5387 , 0.5613]

                    = [53.87% , 56.13%]

Therefore, 90% confidence interval for the true mean % cacao is [53.87% , 56.13%].