# Formula Sheet for Final Exam Plus Tables

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1 Formula Sheet for the Final Exam . If X has a Binomial Distribution with probability of success p and n number of trials, then: 1. 𝑃(? = ?) = ( ? ? ) ? ? ? ?−? , ? = 0, 1, 2, ... , ?, 2. The mean is 𝐸(?) = ?. ? 3. The variance 𝜎2 = ?. ?. ? where ? = (1 − ?) 4. Standard deviation q p n . . Least Square Line: __ 1 __ 0 1 1 0 ^ , x b y b and S S r b where x b b y x y , Any random variable X with normal distribution can be: Standardized to obtain X Z , with new mean = 0 and Variance = 1. For the population proportion p , confidence interval is ˆ p /2 ˆ ˆ / z pq n . The sample size n is large when ?𝜋 > 10 ??? ?(1 − 𝜋) > 10. 1 0.80 0.85 0.90 0.95 0.99 / 2 z 1.28 1.44 1.645 1.96 2.575 In testing 0 H : p = 𝜋 0 , the test statistic is ? = ?̂ − 𝜋 0 𝜋 0 (1 − 𝜋 0 ) ? , ?ℎ??? ?̂ = ? ? The mean of the sampling distribution of the sample mean is 𝜇 ? ̅ = 𝜇 The standard error of the sampling mean ? ̅ is 𝜎 √? Testing the difference between two proportions, Under the null hypothesis: ? 0 ∶ 𝜋 1 − π 2 = 0 The pooled proportion is given by: 2 1 2 1 ˆ n n x x p Pooled The test statistic is 2 1 2 1 2 1 2 1 0 ˆ ˆ ˆ ˆ ˆ ˆ ) ˆ ˆ ( ˆ ˆ n q p n q p p p p p SE p p z Pooled Pooled Pooled Pooled Pooled 2 2 2 1 1 1 2 2 1 ˆ 1 ˆ ˆ 1 ˆ ˆ ˆ Pr % 100 ) 1 ( n p p n p p z p p oportions Population Two between Difference The for Interval Confidence 2 2 / 2 2 / ) 25 (. then, , p about nothing know you If ) ˆ 1 ( ˆ then , experience pass from p of estimate good a know you If E z n E z p p n n p p z p oportion Population for Interval Confidence ) ˆ 1 ( ˆ ˆ : Pr % 100 ) 1 ( 2
2 Population Mean: Sample Mean: 1 1 n i i x x n Range = R = Largest Data Value - Smallest Data Value Population Variance: Sample Variance: 2 2 1 1 ( ) 1 n i i s x x n Population Standard Deviation: Sample Standard Deviation: The Population z-Score: ? = ?−𝜇 𝜎 , and the Sample z-score: ? = ?−? ̅ 𝜎 Interquartile Range IQR=Q 3 -Q 1 ??? 𝑷𝒆?𝒄𝒆????𝒆 ???????? = ? ??? (? + ?) Lower Fence = Q 1 - 1.5 (IQR) and the Upper Fence = Q 3 + 1.5 (IQR) y x n i i i y y x x S S n y y x x r S n SS S n SS Where ) 1 ( ) )( ( ) 1 ( ) 1 ( : 1 2 2 Chi-Square: Goodness of fit Degrees of Freedom (DF) = Number of Categories - 1 Goodness of fit Expected = Percentage * Total count. Independence test or Homogeneity test DF = (Rows - 1)(Columns - 1) ???????????? ?? ????????𝑖?? 𝐸??????? = 𝑇??𝑎? ??? ∗ 𝑇??𝑎? ?????? 𝐺?𝑎?? 𝑇??𝑎? The mean of the sampling distribution of the sample mean is 𝜇 ? ̅ = 𝜇 The standard error of the sampling mean ? ̅ is 𝜎 √? Total Variation = The equation of a linear regression model is: ? ̂ = ? 1 ? + ? 0 , ?ℎ??? 𝑇ℎ? ????? ?? ?ℎ? ???????𝑖?? ?𝑖?? 𝑖? ?𝑖??? ??: ? 1 = ? ? ? ? ? ??? 𝑇ℎ? ? − 𝑖???????? 𝑖?: ? 0 = ? ̅ − ? 1 frequency relative frequency sum of all frequencies 1 2 N x x x N 1 2 n x x x x n 2 s s i i i y y e ˆ 2 2 Observed Expected Expected
3 (1 − 𝛼)100% 𝐶???𝑖????? ???????? ??? ?ℎ? ???????𝑖?? ????: ?̅ ± ? ? √? Testing hypothesis for one mean with sigma is unknown The pooled t -statistic is computed by finding a weighted average of the sample variances and using this average in the computation of the test statistic. The statistic t has degrees of freedom: (n 1 -1) + (n 2 - 1) = n 1 +n 2 -2 ) 1 ( ) 1 ( ) 1 ( ) 1 ( 2 1 2 2 2 2 1 1 2 n n s n s n s Pooled 2 1 2 2 1 2 2 1 1 1 ) ( n n s n s n s y y SE Pooled Pooled Pooled Pooled 2 1 2 1 2 1 0 1 1 n n s y y t Pooled A simple random sample of size n 1 is taken from a population with unknown mean 1 and unknown standard deviation 1 . Also, a simple random sample of size n 2 is taken from a population with unknown mean 2 and unknown standard deviation 2 . If the two populations are normally distributed with common standard deviation, a (1- ) 100% confidence interval about 1 - 2 is given by Lower bound: Upper bound: Constructing a (1- ) 100% Confidence Interval for the Difference of Two Means with equal variances assumption 2 2 1 2 2 2 1 n s n s t y y Pooled Pooled 2 2 1 2 2 2 1 n s n s t y y Pooled Pooled The statistic t has degrees of freedom: (n 1 -1) + (n 2 - 1) = n 1 +n 2 -2 Suppose that a simple random sample of size n 1 is taken from a population with unknown mean 1 and unknown standard deviation 1 . In addition, a simple random sample of size n 2 is taken from a population with unknown mean 2 and unknown standard deviation 2. If the two populations are normally distributed or approximately normal, then approximately follows Student's t -distribution with the degrees of freedom given in the Formula where is the sample mean and s i is the sample standard deviation from population i . Sampling Distribution of the Difference of Two Means: Independent Samples with Population Standard Deviations Unknown (Welch's t ) 2 2 2 1 2 1 2 1 2 1 n s n s y y t i y df s 1 2 n 1 s 2 2 n 2 2 s 1 2 n 1 2 n 1 1 s 2 2 n 2 2 n 2 1 A simple random sample of size n 1 is taken from a population with unknown mean 1 and unknown standard deviation 1 . Also, a simple random sample of size n 2 is taken from a population with unknown mean 2 and unknown standard deviation 2 . If the two populations are normally distributed, a (1- ) 100% confidence interval about 1 - 2 is given by Lower bound: Upper bound: Constructing a (1- ) 100% Confidence Interval for the Difference of Two Means 2 2 2 1 2 1 2 2 1 n s n s t y y 2 2 2 1 2 1 2 2 1 n s n s t y y Paired t-test: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways, where d is the population mean difference of the matched-pairs data. Compute the test statistic which approximately follows Student's t -distribution with n -1 degrees of freedom. The values of d-bar and s d are the mean and standard deviation of the differenced data. n s d t d 0 0 A (1- ) 100% confidence interval for d is given by Lower bound: Upper bound: The critical value t /2 is determined using n -1 degrees of freedom. Confidence Interval for Matched-Pairs Data d t 2 s d n d t 2 s d n n s y t 0