How to find the number of observations needed to discard the null hypothesis with a given probability when the sample size is low?

Question

So I have the following values:

$X_i ~ N(\mu=9.6, \sigma=1.6)$

Samples = 8.4, 7.8, 10.3, 8.7, 10.7, 10.0, 8.8, 12.1

X̄ = 9.6

α = 0.05

I should also mentioned that I have the following hypothesis test:

$H_0: μ \le 8$

$H_1: μ > 8$

I want to find the number of observations needed in my sample so that the probability of discarding the null hypothesis is 0.99.

All of the examples I find online use a Z-table. My problem is however, since the number of samples < 30, shouldn't a T-table be used? But how can you find n with a T-table when the T value depends on n. If I wanted to do some other calculations with with Y, say a hypothesis test to find out if the mean is larger than a given number, I would most definitely use a T-distribution, right? But should I just a Z-table for this specific use-case?

Thank you for the help.

Answer 1

In testing $H_0: \mu \le 8$ against $H_a: \mu > 8$ based on normal data with $\sigma = 1.6,$ are you asking what $n$ is necessary to give power .99 against the specific alternative value $\mu_a = 9.6?$ That is, the power against an alternative that is one SD away from the hypothetical mean $\mu_0 = 8?$

If so, here are results from Minitab statistical software:

Power and Sample Size 

1-Sample Z Test

Testing mean = null (versus > null)
Calculating power for mean = null + difference
α = 0.05  Assumed standard deviation = 1.6


            Sample  Target
Difference    Size   Power  Actual Power
       1.6      16    0.99      0.990742

In such circumstances, power is a function of the distance $\delta = \mu_a - \mu_0.$ Here is Minitab's power curve. It shows increasing power for $n = 16$ as $\delta$ increases; power reaches about .99 for $\delta = 1.6.$

To solve this without specialized software, it might help to find the 'Acceptance region' in terms of the sample mean of $n$ observations when $\mu_0 = 8,\, \alpha = .05,$ and $\sigma = 1.6.$ Then standardize the endpoints of that interval in terms of $\mu = 9.6,$ find the the probability of the interval, subtract from 1 to get an expression for the probability of the 'rejection' region, and solve for $n.$

Note: The statistics text by Ott and Longnecker ('Key Formulas List' at the end of Chapter 5) reduces this process to an equation for $n:$

$$n = \frac{\sigma^2}{\delta^2}(z_\alpha + z_\beta)^2,$$ where $z_\alpha$ cuts 5% of the probability from the upper tail of a standard normal distribution (for a test a level $\alpha$ and $z_\beta$ cuts 1% (fort Type II error 1% and power 99%). In your case, $z_\alpha = 1.645,\, z_\beta = 2.326,$ and $n = 15.77,$ which Minitab rounds up to 16.

Answer 2

It depends whether your test was designed under the assumption that the variance is known or unknown. If unknown you need to use $T$ to find the rejection region for $\alpha =0.05$. If known (and presumably $1.6$) then you should use Z.

Either way, once you have the rejection region, no further use of the T distribution is necessary. You need to calculate the probability that the sample mean, which has distribution $N(9.6,1.6/n)$, falls in the rejection region, and this only requires the normal distribution.