have to guess the age of a random person about whom you have no information you

# Have to guess the age of a random person about whom

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have to guess the age of a random person about whom you have no information, you may guess an interval (0,120) which would probably contain that person’s age. In a confidence interval, this is obtained by having a larger half interval. Since k (Z in a normal distribution), s and n are the three variables which give us the half length, and the values of s and n are fixed, we can only change k. Having a larger k will give us a larger interval. This can be done by increasing the confidence level but it will not be practical for we want to keep the confidence level fixed too. That means we must use another probability distribution which looks like the normal distribution but has a larger variance. Luckily, we have such a distribution, known as Student’s T distribution . The distribution is named after the nickname of William Sealy Gosset who worked for Guinness (Dublin, Ireland) at the time and did not want to use his real name for proprietary restrictions (or so it is rumored). It is also known as T distribution in short and has a bell shape like the normal distribution. T distribution has different variances for different sample sizes, and eventually converges to normal distribution for very large sample sizes. A T table, unlike the Z table, does not give probabilities for every possible T value, but rather lists T values for popular probabilities (like 0.90, 0.95, 0.99, etc in the middle). Still it is possible to find every probability (or T value) by integrating the probability distribution function (or use calculators or programs or websites which do so). T tables list these T values depending on the degrees of freedom which is (n – 1) , therefore T tables typically list degrees of freedom from 1 to 30
after. In practice more degrees o and the standard normal distribu probability (which correspond to Sample size (n) 1 T value 12.706 As expected, for very small samp they get smaller as sample size g Below is the graph of the t distrib Let us now solve the previous ex test statistic become: g2020 = So the test statistic is now; of freedom are seldom required since Central Li ution is utilized. Below you will find some T valu o a Z = ± 1.96). 5 10 20 30 120 2.571 2.228 2.086 2.042 1.98 ples the k values (now shown as t instead of a z gets larger. And about n = 121, t and z are very bution and the general view of a regular t table. xample for a small sample of size n = 19. The con = g1850 g3364 ±g1872 (g2869g2879 g3080 g2870 ) g1871 uni221Ag1866 g1853g1866g1856 g1872 g3030g3028g3039 = g1850 g3364 − g2020 g1871 uni221Ag1866 22 imit Theorem applies ues for a 0.95 middle uni221E 1.96 ) are very large, and y close to each other. nfidence interval and
23 g1852 g3030g3028g3039 = g1850 g3364 − g2020 g1871 uni221Ag1866 = 489− 500 25.3 uni221A19 = −11 5.8 = −1.9 To conclude we will need the t tab from the table. T tables typically cover the probability from −∞ to a positive t value. So we look up (for our two sided test) t (1-α/2),(n-1) = t 0.975,18 = ± 2.101. Since the test statistic is smaller than the critical value, we fail to reject the null hypothesis (μ = 500) at a 95% confidence level. Obtaining a 95% confidence interval for μ is left as an exercise.

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