Respuesta :
Answer:
The 95% confidence interval for the mean weight of all the bags of oranges in the shipment is between 3242.8g and 3357.2g
Step-by-step explanation:
We have the standard deviation of the sample, so we use the t-distribution to solve this question.
T interval
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 10 - 1 = 9
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.95}{2} = 0.975[/tex]. So we have T = 2.262
The margin of error is:
[tex]M = T\frac{s}{\sqrt{n}} = 2.262\frac{80}{\sqrt{10}} = 57.2[/tex]
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 3300 - 57.2 = 3242.8g
The upper end of the interval is the sample mean added to M. So it is 3300 + 57.2 = 3357.2g
The 95% confidence interval for the mean weight of all the bags of oranges in the shipment is between 3242.8g and 3357.2g
