You measure 46 watermelons' weights, and find they have a mean weight of 60 ounces. Assume the population standard deviation is 8.5 ounces. Based on this, construct a 99% confidence interval for the true population mean watermelon weight. Give your answers as decimals, to two places.

A 99% confidence for the true population mean watermelon weight is [56.77 ounces, 63.23 ounces] .

Step-by-step explanation:

We are given that you measure 46 watermelons' weights, and find they have a mean weight of 60 ounces.

Assume the population standard deviation is 8.5 ounces.

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

                             P.Q.  =    ~  N(0,1)    

where, = sample mean weight = 60 ounces

              = population standard deviation = 8.5 ounces

             n = sample of watermelons = 46

              = population mean watermelon weight

Here for constructing a 99% confidence interval we have used a One-sample z-test statistics because we know about the population standard deviation.

So, 99% confidence interval for the population mean, is ;

P(-2.58 < N(0,1) < 2.58) = 0.99  {As the critical value of z at 0.5% level  

                                                     of significance are -2.58 & 2.58}  

P(-2.58 < < 2.58) = 0.99

P( < < ) = 0.99

P( < < ) = 0.99

99% confidence interval for = [ , ]

                                            = [ , ]

                                            = [56.77 ounces, 63.23 ounces]

Therefore, a 99% confidence for the true population mean watermelon weight is [56.77 ounces, 63.23 ounces] .

yor answer is 7/6 times 5

step-by-step explanation:

y=mx+b

step-by-step explanation:

m is slope, b is y-intercept