<u>Answer with explanation:</u>
As per given , we have
A sample of 50 gribbles finds an average length of 3.1 mm with a standard deviation of 0.72.
i.e. n= 50
Degree of freedom = n-1=49
The point estimate of the population mean is equals to the sample mean.
i.e. The best estimate for the length of gribbles= 3.1 mm
Also, population standard deviation is unknown , then the confidence interval would be :
For df= 49 and significance level of 0.05 , the t- value (using t-distribution table) would be :
Now, Margin of error :
95% confidence interval :