Answer: 2.68
Step-by-step explanation:
Given : Sample size : [tex]n=34[/tex]
Mean : [tex]\mu=55\text{ ounces}[/tex]
Standard deviation : [tex]\sigma = 9.5\text{ ounces}[/tex]
Significance level : [tex]\alpha=1-0.9=0.1[/tex]
Critical value : [tex]z_{\alpha/2}=1.645[/tex]
Formula for margin of error :-
[tex]z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\\\\=(1.645)\dfrac{9.5}{\sqrt{34}}\\\\=2.68009413931\approx2.68[/tex]
Hence, the maximal margin of error associated with a 90% confidence interval for the true population mean turtle weight. =2.68
