Question

1. The sample mean weight of 100 chimpanzees was measured to be 82.4 pounds, with a standard deviation of 12.6 pounds. Using a 1% level of significance, what could we claim about the population mean weight for all chimpanzees so that it will fail to be rejected?​

Answer 1

Using the t-distribution, as we have the standard deviation for the sample, we could claim that the mean weight in pounds is in the interval (79.1, 85.7).

What is a t-distribution confidence interval?

The confidence interval is:

$\overline{x} \pm t\frac{s}{\sqrt{n}}$

In which:

• $\overline{x}$ is the sample mean.

• t is the critical value.

• n is the sample size.

• s is the standard deviation for the sample.

The critical value, using a t-distribution calculator, for a two-tailed 99% confidence interval, with 100 - 1 = 99 df, is t = 2.6259.

The other parameters have values given by:

$\overline{x} = 82.4, s = 12.6, n = 100$.

We could claim that the mean is any value in the 99% confidence interval, which has bounds given by:

$\overline{x} - t\frac{s}{\sqrt{n}} = 82.4 - 2.6259\frac{12.6}{\sqrt{100}} = 79.1$

$\overline{x} - t\frac{s}{\sqrt{n}} = 82.4 + 2.6259\frac{12.6}{\sqrt{100}} = 85.7$

More can be learned about the t-distribution at brainly.com/question/16162795