Z-Distribution Articles - Wyzant Tutor Blogshttps://www.wyzant.com/resources/blogs/z_distributionThis is an aggregate of all of the Z-Distribution articles in Wyzant.com's Tutors' Blogs. Wyzant.com is your source for tutors and students.Sun, 26 Jan 2020 06:08:34 -0600https://www.wyzant.com/images/logos/wyzant-logo.pngZ-Distribution Articles - Wyzant Tutor Blogshttps://www.wyzant.com/resources/blogs/z_distributionhttps://www.wyzant.com/resources/blogs/z_distribution725538https://www.wyzant.com/resources/blogs/725538/should_i_use_a_t_or_a_z_distributionYves S.https://www.wyzant.com/resources/users/view/86747260Should I use a t- or a Z-Distribution?<div>We all heard about the simplification to use a t-distribution when sample size is small (n<30) and Z-distribution when n is large (n>30); indeed, with increasing n, the distribution of the sample <span style="text-decoration: underline;">means</span> will converge to the normal distribution thanks to the CLT. But what truly happens when n is not large enough?<br /><br /><span style="text-decoration: underline;">The requirements for a Z-distribution are:</span><br />1) The sample mean is normally distributed <br />AND<br />2) The population standard deviation σ is known (in order to use σ for our statistic test)<br /><br /><br /><span style="text-decoration: underline;">The requirements for a t-distribution are:</span><br />1) The sample size is large (n>30)<br />OR<br />2) The sample is normally distributed (may require separate normality tests)<br />OR<br />3) The population is known to be normally distributed but with standard deviation σ unknown (so we must use s instead).</div>
<div><br />So really, when the sample size is small (n<30), AND when the population from which the sample is taken is known to be normally distributed with a known standard deviation σ, we CAN use the Z distribution (the sample mean from any sample size taken from a normally distributed will be normally distributed). This satisfies condition 1 and 2 for a Z distribution.<br />In case of doubt, you should err on the side of caution and use the t-distribution...</div>
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<div>So what about proportions? Well, here the key is that the <span style="text-decoration: underline;">binomial</span> distribution approximates the normal distribution "under certain conditions":</div>
<div>With n= sample size, p= population probability (and q = 1-p), each individual trial output can only assume two values (1/0, TRUE/FALSE, YES/NO): this is a binary output and follows the Bernoulli distribution.</div>
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<div><span style="text-decoration: underline;">The requirements for using the normal approximation to the binomial distribution are:</span></div>
<div>1) np>5 AND nq>5</div>
<div>AND</div>
<div>2) p is not too close to 0 nor to 1</div>
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<div>When these conditions are not satisfied, we can revert to the binomial distribution (when p = constant, i.e. <span style="text-decoration: underline;">with</span> replacement) or to the hyper-geometric distribution (when p varies on each trial, i.e. <span style="text-decoration: underline;">without</span> replacement) to calculate the probability of K events happening out of N trials given an assumed population probability p (<em>additional arguments k,n for the HG distribution needed</em>).</div>
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<div> </div>Tue, 19 Nov 2019 12:31:26 -06002019-11-19T12:31:26-06:00