1. ## Basic Hypothesis Testing

A sample of 10 seeds are tested to see if the proportion of seeds that germinates one winter has reduced from the usual value of 0.85. Find the critical region for a one-tailed test using a 5% significance level.

Answer: $x\leq6$

2. since n=10 is small you must use the binomial distribution and not approximate this via a normal

$H_0=.85" alt="H_0=.85" /> vs $H_a<.85" alt="H_a<.85" />

I doubt we will get exactly .05

$\alpha=P(X\le c)$ where X is a binomial rv with n=10 and p=.85

DARN close............. $P(X\le 6)\approx$ .049969798878515
at http://stattrek.com/Tables/Binomial.aspx

and u still have an inferior duck

3. Thanks for the reply, I had posted this before seeing it.

Another which I don't understand:

'My research shows that 3 out of 10 children say their favourite colour is red' announced the professor but Miss Smith believed that the proportion was much higher. She asked 6 students, 3 of whom had red as their favourite. She uses a 5% significance level.

a) Test Miss Smith's belief on the basis of this sample (a: He belief at the 5% significance level is unfounded.)
b) She decides to use a larger sample of 20 students. Find how many must choose red as their favourite colour for Miss Smith to have a significant result. (answer: 12)

Edit: I'll get round to updating the avatar soon...

4. $H_0=.3" alt="H_0=.3" /> vs $H_a>.3" alt="H_a>.3" />

Here Our rv X is binomial with n=6 and we assume p=.3 to obtain alpha.

$\alpha=P(X\ge c)$

If she observes 3 I would calculate the p-value as

$P(X\ge 3)\approx .25569$ which is not significant at alpha equal to .05.

As for n=20 I still would use the binomial and not approximate this with a normal.
and I don't see x=12 as the answer here...

If X is Bin(n=20, p=.3), then $P(X\ge 12)\approx 0.00513816153512103$