# Calculating probability of tye 1 error

• Feb 12th 2009, 11:52 AM
Ibn Abi Talib
Calculating probability of tye 1 error
Hi,
im new(Hi)

basically a question says
a sample of 36 parachutes were tested for quality control purpose and their opening altitude is recorded. a statician analysing the data wishes to test the null hypothesis http://thestudentroom.co.uk/latexren...a6a9c75db1.png against http://thestudentroom.co.uk/latexren...ea81d02ad1.png. He dicides on the following criteria:
Accept http://thestudentroom.co.uk/latexren...228769e82a.png if the sample mean is less than or equal to 210m
Reject http://thestudentroom.co.uk/latexren...228769e82a.png if the sample mean is more that 210m

(i) find the probability of making a type I error

can some one help me out please? i thought the prob of type I error was = significance level, but that doesn't seem to be given.

oh yh, mean is 200 and standard deviation is 30
• Feb 13th 2009, 03:55 AM
mr fantastic
Quote:

Originally Posted by Ibn Abi Talib
Hi,
im new(Hi)

basically a question says
a sample of 36 parachutes were tested for quality control purpose and their opening altitude is recorded. a statician analysing the data wishes to test the null hypothesis http://thestudentroom.co.uk/latexren...a6a9c75db1.png against http://thestudentroom.co.uk/latexren...ea81d02ad1.png. He dicides on the following criteria:
Accept http://thestudentroom.co.uk/latexren...228769e82a.png if the sample mean is less than or equal to 210m
Reject http://thestudentroom.co.uk/latexren...228769e82a.png if the sample mean is more that 210m

(i) find the probability of making a type I error

can some one help me out please? i thought the prob of type I error was = significance level, but that doesn't seem to be given.

oh yh, mean is 200 and standard deviation is 30

Calculate $\Pr(\text{Reject} \, H_0 \, | \, H_0 \, \text{true}) = \Pr \left(\overline{X} > 210 \, | \, \overline{X} {\color{white}\frac{.}{.}}\right.$~ Normal $\left. \left(\mu_{\overline{X}} = 200, \sigma_{\overline{X}} = \frac{30}{\sqrt{36}} = 5\right) \right)$.

Aside: The answer tells you what significance level the statistician is using.