Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is reasonable to assume that the population is approximately normal. 21.88 21.76 22.14 21.63 21.81 22.12 21.97 21.57 21.75 21.96 22.20 21.80

Required:
Construct a 90% confidence interval for the mean weight.

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                         P.Q.  =    ~

where, = sample mean weight = = 21.88 ounces

            s = sample standard deviation =  = 0.201 ounces

            n = sample of boxes = 12

             = population mean weight

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

So, 90% confidence interval for the population mean, is ;

P(-1.796 < < 1.796) = 0.90  {As the critical value of t at 11 degrees of

                                                  freedom are -1.796 & 1.796 with P = 5%}  

P(-1.796 < < 1.796) = 0.90

P( < < ) = 0.90

P( < < ) = 0.90

90% confidence interval for = [ , ]

                                        = [ , ]

                                        = [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

answer:      

step-by-step explanation:

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