Answer with explanation:
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
[tex]\overline{x}=3.1\ mm[/tex]
[tex]s=0.72\ mm[/tex]
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 :
[tex]E=t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}[/tex]
For df= 49 and significance level of 0.05 , the t- value (using t-distribution table) would be :
[tex]t_{0.05/2, 49}=t_{0.025, 49}= 2.010[/tex]
Now, Margin of error : [tex]E=(2.010)\dfrac{0.72}{\sqrt{50}}[/tex]
[tex]=0.204664986747\approx0.20[/tex]
95% confidence interval : [tex](\overline{x}-E,\ \overline{x}+E)[/tex]
[tex]i.e.\ (3.1-0.20, 3.1+0.20)=(2.90,\ 3.30)[/tex]
