Respuesta :
Answer:
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. = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{} n} }[/tex] ~ N(0,1)
where, [tex]\bar X[/tex] = sample mean weight = 60 ounces
[tex]\sigma[/tex] = population standard deviation = 8.5 ounces
n = sample of watermelons = 46
[tex]\mu[/tex] = 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, [tex]\mu[/tex] 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 < [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{} n} }[/tex] < 2.58) = 0.99
P( [tex]-2.58 \times {\frac{\sigma}{\sqrt{} n} }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.58 \times {\frac{\sigma}{\sqrt{} n} }[/tex] ) = 0.99
P( [tex]\bar X-2.58 \times {\frac{\sigma}{\sqrt{} n} }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.58 \times {\frac{\sigma}{\sqrt{} n} }[/tex] ) = 0.99
99% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.58 \times {\frac{\sigma}{\sqrt{} n} }[/tex] , [tex]\bar X+2.58 \times {\frac{\sigma}{\sqrt{} n} }[/tex] ]
= [ [tex]60-2.58 \times {\frac{8.5}{\sqrt{46} } }[/tex] , [tex]60+2.58 \times {\frac{8.5}{\sqrt{46} } }[/tex] ]
= [56.77 ounces, 63.23 ounces]
Therefore, a 99% confidence for the true population mean watermelon weight is [56.77 ounces, 63.23 ounces] .
