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I have written:

A predetermined level of significance ($\alpha$ – level) was set to assess the null hypothesis based on a probability value ($p$ value). If the $p$-value is greater than the $\alpha$-level, then the null hypothesis can be accepted. If the $p$-value is less than or equal to the $\alpha$-level then the null hypothesis is rejected. For this analysis the $\alpha$-level was set at $0.025$ giving a $95%$ confidence of the result.

For the $\alpha$-level of $0.025$, it is $95%$ certain that the true difference between population mean values lies within the confidence interval. The estimate for difference is the difference between the population mean values as calculated from the sample data.

The $p$ value was calculated as shown:

$$\displaystyle p = \frac{\bar{x_1} - \bar{x_2}}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} $$

using $n-2$ degrees of freedom.

Is the terminology in that correct? Have I used "$p$ values" correctly there or should it be $t$ value? I'm pretty sure my formula is correct as I can't justify equal variances for both samples. Also, is my conclusion to my hypothesis test correct?

Also, am I right in comparing my $p$ values to my $\alpha$ values or should that say something else?

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    $\begingroup$ An alpha level of 0.025 would correspond to a 97.5% confidence interval -- an alpha of 0.05 would correspond to a 95% confidence interval (i.e. the 0.05 rejection area is split between the two tails of the t-distribution.) $\endgroup$
    – James Stanley
    Commented Apr 16, 2013 at 20:06
  • $\begingroup$ @JamesStanley Would I be able to compare my test statistic to "alpha values from the table" or would I need to reword this to something along the lines of "I can get criticial value corresponding to my alpha value.." and then talk about whether to accept or reject my alternative hypothesis from that? $\endgroup$
    – Kaish
    Commented Apr 17, 2013 at 8:19
  • $\begingroup$ yes, you should be able to check your t-statistic (with n - 2) degrees of freedom with a critical value from a table (with a two-sided alpha of 0.05) to see whether the p-value is < 0.05. $\endgroup$
    – James Stanley
    Commented Apr 17, 2013 at 21:11

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I wouldn't say that you accept the null hypothesis so much as you fail to reject it. Your equation should read z= not p=; that is the equation for a test statistic. Your p-value is the probability that your test statistic takes on values at least as extreme as your observed value. See here for testing difference in means with unequal variances. http://en.wikipedia.org/wiki/Welch%27s_t_test. So for a two-sided alternative test, you could calculate your p-value as 2P(Tv > w) where w is the test statistic and v is degrees of freedom. The function t.test in r can do this for you.

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  • $\begingroup$ So where I have mentioned $p$ values in my text, not the formula, is that correct or should that be $z$ values aswell? $\endgroup$
    – Kaish
    Commented Apr 16, 2013 at 23:30
  • $\begingroup$ You probably wouldn't want to use that formula to test the difference in means because you don't know your population variances and will be using sample variances that may not be equal so you could use the satterthwaite-welch t-test at the link above. That said, if you want to reject/fail to reject your null based on a p-value, then I don't think you would need to mention your test statistic in your decision to reject/fail to reject. I wouldn't change your p-value mentions in your text to test statistic mentions if you aren't going to reject based on critical value. Hope that helps. $\endgroup$
    – user27008
    Commented Apr 17, 2013 at 14:27

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