In a question, when do you calculate t-cal and use to define the acceptance and when do use the sample mean given? I need some clarification plz.

Sorry Dude… don’t understand your question…

Sk… My question is, in trying to define acceptance range, when do you calculate the t-cal ( stastic) and use it with t-critical and when do you use sample mean, which is usually given in the question and use it with t-critical. Hope am clear this time…if not, I will send you a question. Thanks

Post a question and what difference you are observing. The above is quite vague and looking at what you are asking, I am not able to provide you an answer. you could email me at cpkrish@yahoo.com

Here you fellas, A portfolio manager with Aggressive Growth, Inc. asserts that portfolios managed by the firm will outperform the S&P 500 stock index by 5% per year on average. To test this assertion, 51 of the firm’s portfolios are sampled, and on average they outperform the market by 6.3% with a standard deviation of 2.1% in the sample. Using the appropriate test, perform a hypothesis test of the manager’s claim. State the minimum and maximum values of the acceptance range and whether you would accept or reject the manager’s claim at the 1% level of significance Answer: The range is 4.24 to 5.76. Manager’s claim: rejected. Explanation: Sample average of 6.3% lies on the high side of the acceptance range so the null hypothesis that the average fund outperform the index by only 5.o % is rejected. Please breakdown this explanation. Of course calculating the Z cal is not needed here…why?

two ways to do this: Method 1: t-calc = (6.3 - 5) / (2.1/sqrt(50)) = 1.3 / 0.29698484809834996024835463208404 = 4.377 t-crit = 2.678 since t-calc > t-crit --> Manager’s claim is rejected. Other way: t-crit = 2.678 Upper Limit = 5 + 2.678 * 0.29698484809834996024835463208404 = 5.79 Lower Limit = 5 - 2.678 * 0.29698484809834996024835463208404 = 4.21 2.678 is approx # to use – I have used the # for 50 sample size instead of 51 sample size here. Still your value of 6.3 is > 5.79. So still reject the Manager’s claim.

Thank you CP…but why are we using 6.3 in the second solution"way". and not the t-cal. to make the decision. Also this is one tail test…so your t-critical is slighly wrong. It should be 2.403 but it gives almost the same answer, in the first solution…the second solution which is the one given in passmaster…I still can’t get it, why we are using 6.3 to define the acceptance range…I found a few questions like this, with this same way of solving it…so I would like your insight plz…tks

there are two ways to solve such problems. Method 1: Use the Sample Observation (6.3) - Analyst expected (5)/ std dev to calculate T-Calc or Z-Calc compare against the T-Crit / Z-Crit for the issue at hand. Method 2: Use the Analyst Expected +/- T-Crit * Std Deviation (5 +/- …) And then check if Sample Observation lies in that range or outside. (6.3) in either case, if you look at it, you are essentially doing the T-Calc comparison. 6.3 is outside the range because the T-Calc is greater than the T-Crit. Everything else is identical in the two methods.

Thank you CP…I think I will stick to method 1 which seems consistent and easier for me. I am missing the point as to why I am considering 6.3 ( the sample mean) in relation to the range and not the t-cal in relation to range, in method 2. Because if I use 6.3 than it is reject the claim. If I use t-cal it is accept the claim. When are the clues as to when to use the 6.3 ( sample mean) in making the claim’s decision…I feel am close to getting it but missing a link or a concept…even now that am wake oh boy!! All the same…I really appreciate your help. I will do more practise questions and hope to nail it. Cheers! though…

CP… I think I got it. There are 2 methods…the Null and alternative hypothesis method and t-statistic. So in the Null and alternative hypothesis, is where the sample mean is used in relation to the range which is calculated using the claim value. And of course the t-statistic method. Thank you much CP… I think I have it now…Wow!! F

In this question why would we not use the Z-score to find our critical value. Sample size is 51. Isn’t it acceptable to be using Z-test statistic?

If you have different answers for the T and the Z, you should probably go with the T (not reject) result. Both are approximate tests here, so you want to be conservative.

If population variance is not known, then you use the t-statistic (preferrably). Just as Joey stated above.