Width of two sided hypothesis test

Question

I'm not sure how to find the width for this. I know how to find the confidence interval had they given the sample mean, but in this case I'm not sure how I would find the width.

Any help would be much appreciated.

Answer 1

I'm not doing the calculation but here's how you would go about it.

You know that the confidence interval is of the form $(\mu - e,\mu + e)$ where $e$ is the error. You know how to calculate $e$ by finding the $z$ or $t$ score for the 98% confidence interval (1% on either side). Multiply the standard error by the z/t score. Now you have $e$. And the width of the distribution is just $2e$.

Hope that helps

Answer 2

The confidence interval is $$\left[\overline x-t_{(1-\frac{\alpha}{2},n-1)}\cdot \frac{s}{\sqrt n}, \ \overline x+t_{(1-\frac{\alpha}{2},n-1)}\cdot \frac{s}{\sqrt n} \right]$$

Thus the width is

$$\overline x+t_{(1-\frac{\alpha}{2},n-1)}\cdot \frac{s}{\sqrt n}-\left( \overline x-t_{(1-\frac{\alpha}{2},n-1)}\cdot \frac{s}{\sqrt n} \right)=2\cdot t_{(1-\frac{\alpha}{2},n-1)}\cdot \frac{s}{\sqrt n} $$

where $s$ is square root of the sample variance, $\alpha=0.02$ is the significance level and $t_{(1-\frac{\alpha}{2},n-1)}$ is the value of the cdf of the t-distribution.