# not sure which on to choose

• Jun 6th 2008, 04:56 AM
crashuk
not sure which on to choose
) A tyre manufacturer claims that his new steel-belted tire has a mean life expectancy of 40,000 miles. A consumer association decides to test these claims against the alternatives that the mean life expectancy is less than 40,000 miles. 100 tyres are selected at random and the sample mean is found to be 38,500 miles with the sample standard deviation of 2,000 miles.

Perform the hypothesis test at the 5% (or 0.05) level of significance.

because its a sample standard deviation i should use xbar- meu = a and then SD/ n square root= b
so is it 38500- 40000 = 1500 and the 2000 / sq of 100 = 10 = 2000/ 10 = 200 so
1500/ 200 = 7.5

so
h0 meu= 40000
h1 meu doesnt=40000
5%
• Jun 6th 2008, 05:07 AM
crashuk
xbar = 30 s squared = 100 and n = 64
Test the hypothesis
q H0 : μX = 25 and
H1 : μX ≠ 25:

would it be 30-25/ s= 10 n= 10=
• Jun 6th 2008, 07:09 AM
mr fantastic
Quote:

Originally Posted by crashuk
) A tyre manufacturer claims that his new steel-belted tire has a mean life expectancy of 40,000 miles. A consumer association decides to test these claims against the alternatives that the mean life expectancy is less than 40,000 miles. 100 tyres are selected at random and the sample mean is found to be 38,500 miles with the sample standard deviation of 2,000 miles.

Perform the hypothesis test at the 5% (or 0.05) level of significance.

because its a sample standard deviation i should use xbar- meu = a and then SD/ n square root= b
so is it 38500- 40000 = 1500 and the 2000 / sq of 100 = 10 = 2000/ 10 = 200 so
1500/ 200 = 7.5

so
h0 meu= 40000
h1 meu doesnt=40000
5%

You need to say what test you're using. I assume you're using a z-test since, although the population sd is unknown, n is large and the assumption of normality is reasonable.

So you have z = -7.5. What's the critical value of z for a one-sided test at the 0.05 level of significance? Is -7.5 larger or smaller than this value? What conclusion do you draw?
• Jun 7th 2008, 06:52 AM
crashuk
i thought it was a two sided test?
= and not= ?
• Jun 7th 2008, 04:01 PM
mr fantastic
Quote:

Originally Posted by crashuk
i thought it was a two sided test?
= and not= ?

" test these claims against the alternatives that the mean life expectancy is less than 40,000 miles."

is the only alternative stated in the question. This implies a one sided test.
• Jun 8th 2008, 09:45 PM
crashuk
Quote:

Originally Posted by mr fantastic
You need to say what test you're using. I assume you're using a z-test since, although the population sd is unknown, n is large and the assumption of normality is reasonable.

So you have z = 7.5. What's the critical value of z for a one-sided test at the 0.05 level of significance? Is 7.5 larger or smaller than this value? What conclusion do you draw?

how do you work that out?
• Jun 8th 2008, 11:17 PM
mr fantastic
Quote:

Originally Posted by crashuk
how do you work that out?

Usually you would use a table of critical values to get the critical value (it can of course be got from the usual four figure tables):

The critical value of z is -1.64.

-7.5 < -1.64 therefore the result is significant at the 0.05 level.

Therefore you reject the null hypothesis.

The manufacturers claim appears to be bogus.
• Jun 9th 2008, 03:34 AM
Nyoxis
Unbiased estimate?
Shouldn't you use an unbiased estimate of the sample variance, because it's not normally distributed? like:

$\frac{n}{n-1}(2000^2) = 4040404.04$

Then if z<-1.6448, reject $H_{o}: \mu=40000$

and z= $\frac{38500-40000}{\sqrt{\frac{4040404.04}{100}}}$

=-7.462405778<-1.6448, so reject $H_{o}$
• Jun 9th 2008, 04:02 AM
mr fantastic
Quote:

Originally Posted by Nyoxis
Shouldn't you use an unbiased estimate of the sample variance, because it's not normally distributed? like:

$\frac{n}{n-1}(2000^2) = 2020.20202$

Then if z<-1.6448, reject $H_{o}: \mu=40000$

and z= $\frac{38500-40000}{\sqrt{\frac{4040404.04}{100}}}$

=-7.462405778<-1.6448, so reject $H_{o}$

I think n = 100 is large enough for the correction factor to be ignored. Also, I think an asusmption of normality is reasonable (the lifetime of a tyre might just follow a Weibull distribution, say. But the value of the parameters would I think make a normal approximation reasonable)