What is the right h0?

Question

the problem is : A manufacture claims that a new brand of air-conditioning unit uses only 6.5 kilowatts of electricity per day,A consumer agency believe the true figure is higher and runs a test on a sample of size 50.if the sample mean is 7.0 kilowatts with a standard deviation of 1.4,should the manufacturer's claim be rejected at a significance level of 5% ?Of 1%?

the answer is :Assuming an SRS we have:

H0:mean=6.5
HA:mean>6.5
(omitted the following process)

here ,H0:mean=6.5 in the book,i don't think it is right,it should be:
H0:mean<=6.5
HA:mean>6.5

am i right?

Answer 1

The null hypothesis is generally done so that you make an assumption that there's "no-difference." When you make the assumption that H0:mean<=6.5, you're injecting into the null hypothesis that the mean kilowatt use is possibly < 6.5. We need more info on what kind of test you're computing, but in general the computations to get your test-statsitic and p-value are done with this "no-difference" assumption and not a composite of "<" involved.

Here is a blog-like article on the null hypothesis: http://udel.edu/~mcdonald/stathyptesting.html

These kinds of problems though are naturally suited in a bayesian environment. There's a great light bulb example in Bayesian Reliability (http://books.google.com/books/about/Bayesian_Reliability.html?id=GIhbskry6NYC ), where a manufacturer makes a claim about light bulb lifetimes. The skeptic students grab some light bulbs test them, and combine the manufacturers claim with their new test data to measure the chances the manufacturer was right.

In the same suite, you could measure how likely it is the mean<=6.5 using the consumers agency data, combined with what the manufacturer claims.

By doing that you've gotten rid of the H0,HA approach. However it's much more complex to do.

Good luck.

Answer 2

In your case I would just use a confidence interval. You say that from a sample of 50 the observed mean is 7 and standard deviation is 1.5.

You can compute a confidence interval as $mean\pm z\frac{sd}{\sqrt{N}}$, where $z$ depends on your confidence level. For a 95% interval, $z=1.96$, for 99% $z=2.58$.

So you would have $7\pm1.96\frac{1.4}{\sqrt{50}}=7\pm0.388=[6.612,7.388]$. Therefore, the consumer agency can claim it to be larger than 6.5.