The t-Distribution and z-Distribution PDF

The t-Distribution The t-Distribution ▪Student’s t-distribution ▪is any member of a family of continuous probability distributions that arises when estimating the mean of a normally distributed population in situations where the sample size is small and the population standard deviation is unknown....

The t-Distribution The t-Distribution ▪Student’s t-distribution ▪is any member of a family of continuous probability distributions that arises when estimating the mean of a normally distributed population in situations where the sample size is small and the population standard deviation is unknown. ▪William Sealy Gosset under the pseudonym Student. The t-Distribution ▪The t-distribution is symmetric and bell-shaped but has heavier tails ▪it is more prone to producing values that fall far from its mean. This can be used to construct a confidence interval for the true mean. ▪The t distribution incorporates the fact that for smaller sample sizes the distribution will be spread out using degree of freedom. Degrees of freedom ▪defined as the number of values or “observations” in the data that are free to vary when estimating statistical parameters. ▪It help us to achieve desired confidence level. For confidence intervals, the degree of freedom will always be df = n – 1, or one less the sample size. ▪The smaller the df, the flatter the shape of the distribution and has a greater area under the tails. Normally, you use t- table when the sample size is small (n < 30) and the population standard deviation σ is unknown. Find the degree of freedom of n=4 and n=7 and illustrate the graph. df = n - 1 df = 4 – 1 =3 df = 7 – 1 =6 Note: Increasing the degree of freedom will decreases the critical t values and get closer to zero to keep the same area under the curve. Using t-Table, find the confidence coefficient of n=12 and 95% confidence and illustrate the graph. Answer: The confidence coefficient = 2. 201 Using t-Table, find the confidence coefficient of n=12 and 95% confidence and illustrate the graph. The Confidence Interval Confidence Interval ▪An interval estimate is a range of values or interval (with lower and upper limits) used to estimate the population parameter. ▪This estimate may or may not contain the true parameter value. The parameter is specified as being between two values. It is usually in the form of a < ϴ < b, which tells that the estimated parameter (ϴ) is between two values ( a and b ) at a certain level of confidence. The t –distribution ▪if σ is unknown and n < 30, use tά/2 𝑥̿ = sample mean s = sample standard deviation n = sample size tα/2 = t-value with n-1 degrees of freedom, that leaves an area of α/2 The t –distribution ▪Margin of error However, when σ is not known (as is often the case), the sample standard deviation s is used to approximate σ. So, the formula for E is modified. Compute the margin of error of the 90% confidence interval estimate of µ when s = 5, n = 16. Given: confidence level = 90% s=5 n = 16 𝑠 𝐸 = 𝑡𝑎 ( ) 2 𝑛 5 𝐸 = 1.753( ) 16 𝐸 = 2.19 Hence, the margin of error is 2.19. The average hemoglobin reading for a sample of 20 teachers was 16 grams per 100 milliliters with a sample standard deviation of 2 grams. Find the 95% confidence interval of the true mean. Given: confidence level = 95% s=2 n = 20 x = 16 Solution: 2 2 16 – 2.093 ( )< μ < 16 + 2.093 ( ) 20 20 16 – 0.94 < μ < 16 + 0.94 15.06 < µ < 16.94 or (15.06 , 16.94) The z-distribution if σ is known and n ≥ 30, use zά/2 where: x = sample mean σ = population standard deviation n = sample size zά/2 = z value that leaves an area of α/2 The z –distribution ▪Margin of error 𝜎 E = za/2 ( ) 𝑛 ▪Margin of Error refers to the maximum acceptable difference (determined by α) between the observed sample statistic (mean or proportion) and the true population parameter (mean or proportion). Compute the margin of error of the 95% confidence interval estimate of µ when σ=10, n=50. Given: confidence level = 95% σ = 10 n = 50 Solution: 10 E = 1.96 ( ) 50 E= 2.77 Hence, the margin of error is 2.77.  Compute the 98% confidence interval estimate of µ given the ff: σ=6.4, n=40, and 𝑥̅=42 Solution: σ σ 𝑥̅ – za/2 ( ) < μ < 𝑥̅ + za/2 ( ) 𝑛 𝑛 6.4 6.4 42 – 2.33( ) < μ < 42 + 2.33( ) 40 40 39.64 < μ < 44.36 or or (39.64 , 44.36) ACTIVITY # 1.6 1. Compute the 90% confidence interval estimate of µ given the ff: σ=8.2, n=48, and 𝑥̅=52. Draw the graphical representation for the interval. 2. Compute the margin of error of the 98% confidence interval estimate of µ when σ=5, n=52. 3. Compute the margin of error of the 95% confidence interval estimate of µ when s = 6.3, n = 25. 4. Compute the 90% confidence interval estimate of µ given the ff: s=4, n=23, and 𝑥̅=45. Draw the graphical representation for the interval.