Question

You measure 34 turtles' weights, and find they have a mean weight of 55 ounces. Assume the population standard deviation is 9.5 ounces. Based on this, what is the maximal margin of error associated with a 90% confidence interval for the true population mean turtle weight. Give your answer as a decimal, to two places

Answer 1

Answer: 2.68

Explanation:

Given : Sample size :
`n=34`

Mean :
`\mu=55\text{ ounces}`

Standard deviation :
`\sigma = 9.5\text{ ounces}`

Significance level :
`\alpha=1-0.9=0.1`

Critical value :
`z_(\alpha/2)=1.645`

Formula for margin of error :-

`z_(\alpha/2)(\sigma)/(√(n))\\\\=(1.645)(9.5)/(√(34))\\\\=2.68009413931\approx2.68`

Hence, the maximal margin of error associated with a 90% confidence interval for the true population mean turtle weight. =2.68