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il63 [147K]
2 years ago
15

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?​
Mathematics
1 answer:
tia_tia [17]2 years ago
5 0

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).

<h3>What is a t-distribution confidence interval?</h3>

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 <em>99% confidence interval</em>, with 100 - 1 = <em>99 df</em>, 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

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