# [S2000Magician or Tickersu] Help: Doubt related Hypothesis Testing

For a past few days I have been struggling to understand this concept. I would really appreciate if you could help me out with this. I’ll try to explain my query through an illustration.

Illn: Let our sample mean be 5%, standard error being 0.75%. We are hypothesizing our population mean to be equal to 6.25% . Let’s assume the level of significance to be 5%. Further assuming that z distribution is being followed, since the test is a two tailed test, the z value would be 1.96. Now suppose we are required to solve the question using confidence interval.

The value we will add or less to fit the level of significance is (Std err * z value) ie (1.96*0.75%) ie our add or subtract value would be 1.47.

Now to make a conclusion if null hypothesis is true or not amongst the following which logical reasoning do we make

Option 1. Build confidence intervals around hypothesized population mean.

We will build confidence intervals around hypothesized population mean and check if the sample mean rests within that range. So in the given question we would create a confidence interval around 6.25. Ie range within which sample mean should rest to prove null hypothesis is 6.25-1.47 & 6.25+1.47 ie if sample mean rests between 4.78 & 7.72. In the given case, since the sample mean of 5% does actually rest between 4.78 & 7.72 we fail to reject null.

Option 2. Build confidence intervals around sample mean.

We will build confidence intervals around sample mean and check if the hypothesized population mean rests within that range. So in the given question we would create a confidence interval around 5. Ie range within which hypothesized population mean should rest to prove null hypothesis is 5-1.47 & 5+1.47 ie if hypothesized mean rests between 3.53 & 6.47. In the given case, since the hypothesized mean of 6.25% does actually rest between 3.53 & 6.47 we fail to reject null.

Which of the above is the actual reasoning on basis we accept or reject hypothesis ?. I would also really appreciate if you could assist me on how to build confidence intervals in one tailed tests in following both cases where

1. H0: μ0>=6.25% 2. μ0 <=6.25%

First: thank you for including me in such esteemed company as tickersu. I’m extremely flattered.

Second: you build the confidence interval around the sample mean.

If that’s the case, then why is the formula for test statistic. (Xbar-u0)/ S.E. I mean shouldn’t it be the other way round ie. (u0- xbar)/ S.E. Since test statistic is essentially no of standard deviations around the sample mean and not the other way round.

Also in the given case how would we interpret one tailed hypothesis in the following two cases

Case1. H0: u0>= 6.25

Case2. H0: u0<=6.25

In both of the cases we would get the null hypothesis as true.So I wanted to ask what is actually the case, since both case 1 and case 2 are mutually exclusive.

Forgive me if there was any stupidity involved in my query. Thank you again.

To be blunt, you cannot make any conclusions regarding the null hypothesis being true without risking a Type II error, and in general you’re not able to say the null is true from a simple confidence interval or hypothesis test. You can only find evidence contradicting the null hypothesis, never in support of it with these methods.

You may not conclude the null is true in either case.

Thank you for replying. That seems to be reasonable. Could you also explain the reason why we use (x̅-μ0)/S.E for finding test statistic as opposed to (μ0-x̅)/S.E. The reason I think so is because text statistic indicates how much standard deviations our estimate is away from our mean. Thank you again for replying.

It follows the general convention of observed minus expected, so if we observe larger than expected we get a positive deviation which, I think, is slightly easier to quickly understand the standardized value as opposed to how you’ve written it with a negative value indicating our observation was greater than expected (mu0).

I’m assuming this is more or less the reason.

Thanks a lot.