Question

We would like to estimate the true mean weight of a certain breed of cats. Suppose it is known that weights follow a normal distribution with standard deviation 5.5 pounds.

a. What sample size is required to estimate the true mean weight to within 4 pounds with 94% confidence?

It is calculated that, in order to estimate the true mean weight of a different breed of cats (that is, σ = 5.5 no longer applies) to within 4 pounds with 95% confidence, we would need a sample of 50 cats.

b.How many cats would we need to sample in order to estimate the true mean weight to within 3 pounds with 95% confidence?

Answer 1

Answer on Question #84602 – Math – Statistics and Probability

Question

We would like to estimate the true mean weight of a certain breed of cats. Suppose it is known that weights follow a normal distribution with standard deviation 5.5 pounds.

a. What sample size is required to estimate the true mean weight to within 4 pounds with 94% confidence?

Solution

a. $SE = z_{0.03} \frac{\sigma}{\sqrt{n}} \rightarrow n = \left(\frac{z_{0.03} \sigma}{SE}\right)^2 = \left(\frac{1.881 + 5.5}{4}\right)^2 = 7.$

Question

It is calculated that, in order to estimate the true mean weight of a different breed of cats (that is, $\sigma = 5.5$ no longer applies) to within 4 pounds with 95% confidence, we would need a sample of 50 cats.

b. How many cats would we need to sample in order to estimate the true mean weight to within 3 pounds with 95% confidence?

Solution

b. $\sigma = \frac{SE\sqrt{n}}{z_{0.025}} = \frac{4 + \sqrt{50}}{1.96} = 14.4.$

$$n = \left(\frac{z_{0.025} \sigma}{SE}\right)^2 = \left(\frac{1.96 + 14.4}{3}\right)^2 = 89.$$

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