Question

Find a 75% confidence interval for the population average weight μ of all adult mountain lions in the specified region. (Round your answers to one decimal place.)

Answer 1

Answer:

75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5 pounds.

Step-by-step explanation:

The question is missing. It is as follows:

Adult wild mountain lions (18 months or older) captured and released for the first time in the San Andres Mountains had the following weights (pounds): 69  104  125  129  60  64

Assume that the population of x values has an approximately normal distribution.

Find a 75% confidence interval for the population average weight μ of all adult mountain lions in the specified region. (Round your answers to one decimal place.)

75% Confidence Interval can be calculated using M±ME where

• M is the sample mean weight of the wild mountain lions ($\frac{69 +104 +125 +129+60 +64}{6} =91.8$)

• ME is the margin of error of the mean

And margin of error (ME) of the mean can be calculated using the formula

ME=$\frac{t*s}{\sqrt{N} }$ where

• t is the corresponding statistic in the 75% confidence level and 5 degrees of freedom (1.30)

• s is the standard deviation of the sample(31.4)

• N is the sample size (6)

Thus, ME=$\frac{1.30*31.4}{\sqrt{6} }$ ≈16.66

Then 75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5