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%.

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. =  ~

where, = sample mean % cacao = 55%

             s = sample standard deviation = 4%

             n = sample of pieces = 36

              = 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, is ;

P(-1.6895 < < 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 < < 1.6895) = 0.90

P( < < ) = 0.90

P( < < ) = 0.90

90% confidence interval for = [ , ]

                    = [ , ]

                    = [0.5387 , 0.5613]

                    = [53.87% , 56.13%]

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

answer: idk

step-by-step explanation:

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