# Thread: Samplesof two types of electric components ....State the null hypothesis

1. ## Samplesof two types of electric components ....State the null hypothesis

Question :

Samples of two types of electric component were tested for length of life and following data were determined:

$\displaystyle **\qquad \qquad \qquad \qquad Type1 \qquad \qquad\qquad \qquad Type 2$

$\displaystyle Sample \ size \qquad \qquad n_1 = 8 \qquad \qquad \qquad \qquad n_2 = 7$

$\displaystyle Sample \ mean \qquad \quad \bar{x_1} = 1234 hrs \qquad \qquad \quad \bar{x_2} = 1036 hrs$

$\displaystyle Sample \ SD's \qquad \quad S_1 = 36 hrs \qquad \qquad \qquad S_2 = 40 hrs$

Is the difference in means sufficient to say that Type 1 is superior to Type 2 regarding length of life ? State the null hypothesis.

2. Originally Posted by zorro
Question :

Samples of two types of electric component were tested for length of life and following data were determined:

$\displaystyle **\qquad \qquad \qquad \qquad Type1 \qquad \qquad\qquad \qquad Type 2$

$\displaystyle Sample \ size \qquad \qquad n_1 = 8 \qquad \qquad \qquad \qquad n_2 = 7$

$\displaystyle Sample \ mean \qquad \quad \bar{x_1} = 1234 hrs \qquad \qquad \quad \bar{x_2} = 1036 hrs$

$\displaystyle Sample \ SD's \qquad \quad S_1 = 36 hrs \qquad \qquad \qquad S_2 = 40 hrs$

Is the difference in means sufficient to say that Type 1 is superior to Type 2 regarding length of life ? State the null hypothesis.
The null htpotheseis is that the means are equal, now what test do you use for sample means from small samples with standard deviation estimated from the samples?

CB

3. ## Is this correct?

Originally Posted by CaptainBlack
The null htpotheseis is that the means are equal, now what test do you use for sample means from small samples with standard deviation estimated from the samples?

CB
This is my work please take a look and tell me if i have done any thing wrong

$\displaystyle H_0$ : $\displaystyle \mu - \mu$ = 0
$\displaystyle H_a$ : $\displaystyle \mu - \mu$ > 0
$\displaystyle \alpha$ : 0.05

Is this correct ?

4. Originally Posted by zorro
This is my work please take a look and tell me if i have done any thing wrong

$\displaystyle H_0$ : $\displaystyle \mu - \mu$ = 0
$\displaystyle H_a$ : $\displaystyle \mu - \mu$ > 0
$\displaystyle \alpha$ : 0.05

Is this correct ?
It looks like you are supposed to use a two sample unpaired t-test (because the sample sizes are small) with unequal sample sizes (see Student's t-test - Wikipedia, the free encyclopedia)

The problem with this is that this requires that the distribution of component lives is normal, which is not usually the case and not given in the question.

CB

5. ## Do u mean this test

Originally Posted by CaptainBlack
The null htpotheseis is that the means are equal, now what test do you use for sample means from small samples with standard deviation estimated from the samples?

CB
Do u mean this test

$\displaystyle t$ = $\displaystyle \frac{\bar x_1 - \bar x_2 - k}{s_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

where $\displaystyle s_p$ = $\displaystyle \frac{(n_1 -1)s_1 ^2 + (n_2 -1)s_2 ^2}{n_1 + n_2 - 2}$

but what should be the null hypothesis

6. Originally Posted by zorro
Do u mean this test

$\displaystyle t$ = $\displaystyle \frac{\bar x_1 - \bar x_2 - k}{s_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

where $\displaystyle s_p$ = $\displaystyle \frac{(n_1 -1)s_1 ^2 + (n_2 -1)s_2 ^2}{n_1 + n_2 - 2}$

but what should be the null hypothesis
The null hypothesis is that the means are equal (and so in your notation that $\displaystyle k=0$)

CB

7. ## Is this correct?

Originally Posted by CaptainBlack
The null hypothesis is that the means are equal (and so in your notation that $\displaystyle k=0$)

CB

$\displaystyle H_0 : \mu_1 - \mu_2 = 0$
$\displaystyle H_a : \mu_1 - \mu_2 \ge 0$
$\displaystyle \alpha : 0.05$

Is this correct ?

8. Originally Posted by zorro
$\displaystyle H_0 : \mu_1 - \mu_2 = 0$
$\displaystyle H_a : \mu_1 - \mu_2 \ge 0$
$\displaystyle \alpha : 0.05$

Is this correct ?
Yes, for a one sided test ($\displaystyle H_a$ is that $\displaystyle \mu_1 > \mu_2$, note $\displaystyle >$ not $\displaystyle \ge$ as this includes the null-hypothesis) the two sided test would have $\displaystyle H_a: |\mu_1-\mu_2| > 0$

CB

9. From the sample st deviations, I guess we can assume that the population variances are equal.

Originally Posted by CaptainBlack
Yes, for a one sided test ($\displaystyle H_a$ is that $\displaystyle \mu_1 > \mu_2$, note $\displaystyle >$ not $\displaystyle \ge$ as this includes the null-hypothesis) the two sided test would have $\displaystyle H_a: |\mu_1-\mu_2| > 0$

CB
$\displaystyle H_0 : \mu_1 - \mu_2 = 0$
$\displaystyle H_a : | \mu_1 - \mu_2 | > 0$
$\displaystyle \alpha = 0.05$

- Reject the null hypothesis if t > $\displaystyle t_{\alpha,n_1 + n_2 -2}$ ie t > 1.771

$\displaystyle t = \frac{ \bar x_1 - \bar x_2 - 0}{S_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$
where

$\displaystyle S_p = \frac{(n_1 - 1) s_1 ^2 + (n_2 - 1) s_2 ^2}{n_1 + n_2 -2}$

$\displaystyle S_p$ = 1436

Substituting $\displaystyle S_p$ in the equation weget

t = 0.26 < 1.771

|t| < 1.771 therefore null hypothesis cannot be rejected

Is this correct ?

11. Originally Posted by matheagle
From the sample st deviations, I guess we can assume that the population variances are equal.
There is no reason to assume the variances are equal other than the near coincidence in numerical values. Given the high variability in sample variance (with such small sample sizes) I would stick to treating the variances as unequal.

CB

12. Originally Posted by zorro
$\displaystyle H_0 : \mu_1 - \mu_2 = 0$
$\displaystyle H_a : | \mu_1 - \mu_2 | > 0$
$\displaystyle \alpha = 0.05$

- Reject the null hypothesis if t > $\displaystyle t_{\alpha,n_1 + n_2 -2}$ ie t > 1.771

$\displaystyle t = \frac{ \bar x_1 - \bar x_2 - 0}{S_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$
where

$\displaystyle S_p = \frac{(n_1 - 1) s_1 ^2 + (n_2 - 1) s_2 ^2}{n_1 + n_2 -2}$

$\displaystyle S_p$ = 1436

Substituting $\displaystyle S_p$ in the equation weget

t = 0.26 < 1.771

|t| < 1.771 therefore null hypothesis cannot be rejected

Is this correct ?
There is an error in the equation for pooled standard deviation, looks like you have used the pooled variance.

CB

13. You can use an F test to determine if the population variances are equal.
If you decide they are, then the test stat for the means is

$\displaystyle {\bar x_1- \bar x_2\over s_p\sqrt{ {1\over n_1}+{1\over n_2}}}$

otherwise you should use Satterthwaite's approximation

$\displaystyle {\bar x_1- \bar x_2\over \sqrt{ {s_1^2\over n_1}+{s_2^2\over n_2}}}$

http://www.unm.edu/~marcusj/PSE.pdf
which has a typo in it, they squares the pooled sample variance incorrectly in line 11

14. ## please provide the correct solution

Originally Posted by CaptainBlack
There is an error in the equation for pooled standard deviation, looks like you have used the pooled variance.

CB
Could u please provide me with the right solution. If the formula which i have not used is right then which one should i use

In the prev post i asked u if the formula which i used was correct or no , at that time u didnt pointed out this error.....

please let me know what is the right formula and also why , so that i dont make the same mistake again .

15. Originally Posted by zorro
Could u please provide me with the right solution. If the formula which i have not used is right then which one should i use

In the prev post i asked u if the formula which i used was correct or no , at that time u didnt pointed out this error.....

please let me know what is the right formula and also why , so that i dont make the same mistake again .
What your formula gives is $\displaystyle S_p^2$ not $\displaystyle S_p$, so just take the square root of what you have.

CB

Page 1 of 2 12 Last