Question

From a sample of 10 squirrels the average weight was 511 grams with standard deviation of 160 grams.

What is the t value for a 95% confidence interval?


What are the lower and upper limits of the 95% confidence interval?

Answer 1

Answer on Question #44384 – Math - Statistics and Probability

Task:

From a sample of 10 squirrels the average weight was 511 grams with standard deviation of 160 grams.

What is the t value for a 95% confidence interval?

What are the lower and upper limits of the 95% confidence interval?

Solution:

The confidence interval is:

$$\bar{x} - t \frac{\delta}{\sqrt{n}} < a < \bar{x} + t \frac{\delta}{\sqrt{n}}$$

where $\bar{x} = 511$, $\delta = 160$, $n = 10$. $\Phi(t)$ – Laplace function

$P = 0,95 = 2\Phi(t)$; $\Phi(t) = \frac{0,95}{2} = 0,475$. From the table we can see that $t = 1,96$.

Now we can use the formula:

$$511 - 1,96 \frac{160}{\sqrt{10}} < \alpha < 511 + 1,96 \frac{160}{\sqrt{10}}$$

So, lower and upper limits of the 95% confidence interval are:

$$411,82 < \alpha < 610,18$$

Answer:

$t = 1,96$, $411,82 < \alpha < 610,18$

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