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

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

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. =
`(\bar X-\mu)/((s)/(√(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 <
`(\bar X-\mu)/((s)/(√(n) ) )` < 1.6895) = 0.90

P(
`-1.6895 * {(s)/(√(n) ) }` <
`{\bar X-\mu}` <
`1.6895 * {(s)/(√(n) ) }` ) = 0.90

P(
`\bar X-1.6895 * {(s)/(√(n) ) }` <
`\mu` <
`\bar X+1.6895 * {(s)/(√(n) ) }` ) = 0.90

90% confidence interval for
`\mu` = [
`\bar X-1.6895 * {(s)/(√(n) ) }` ,
`\bar X+1.6895 * {(s)/(√(n) ) }` ]

= [
`0.55-1.6895 * {(0.04)/(√(36) ) }` ,
`0.55+1.6895 * {(0.04)/(√(36) ) }` ]

= [0.5387 , 0.5613]

= [53.87% , 56.13%]

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