Please help me solve this question.

Question

Packets of fish food have weights that are distributed with standard deviation 2.3 g. A random sample of 200 packets is taken. The mean weight of this sample is found to be 99.2 g. Calculate a 99% confidence interval for the population mean weight.

Answer 1

x̄ = 99.2

σ = 2.3

n = 200

Z = 2.576

= 99.2 ± (2.5758 × 2.3/√200) =

= 99.2 ± 0.419

= (98.781 - 99.619)

We can be 99% confident that the population mean of fish packet weights will lie within 98.781 - 99.619

Answer 2

A confidence interval is a RANGE around a descriptive statistic, so is two numbers. A confidence interval is calculated by taking a descriptive statistic (like the mean) and then adding/subtracting the standard error times the a critical value to/from this statistic.

As such, a confidence interval is centered on the descriptive statistic you are making an estimate about. In this case, the mean of 99.2 is where we begin. To construct the interval, you'll now need the standard error of the mean, which in this case is S/√N or 2.3/√200 because N is the sample size (200). Now we need a critical value. To find a critical value, you need the "level of confidence", the degrees of freedom, and a t-table.

The level of confidence here is 99%, the degrees of freedom in this problem are N - 1 or 199. Here is a link to a website that can help you look up the correct critical value:

https://www.graphpad.com/quickcalcs/statratio1/

You choose the "t-ratio" and type in 199 for DF and .01 for the "probability." This is because a 99% confidence interval means we have an alpha = .01 (100% - 99% = 1% or alpha =.01). This will give you the correct critical value.

Now you have all you need to construct the interval, which again will two numbers:

99.2 + ((2.3/√200) * critical value) and then...

99.2 - ((2.3/√200) * critical value)