FINAL EXAM - FALL 2023 Review Sheet- SOLUTION

FINAL EXAM REVIEW - SOLUTION MAT 150.5 Z-test t-test (one-sample) _ M — oy = standard error of the mean sy = estimated standard error p. 198-9 p. 211-2 2 = yari _ s s = Remember: s =variance §,, = ; M \/; Independent t-test Independent-measures or between-subject design Null hypothesis Ho: By - M2 =0 (or py= py) Alternative hypothesis Hiipy - #0 (orp #H, or Py<p, or py>py) =g t-test: double elements of single t-test formula Sm S my—my) Compare mean difference (top) with difference expected by chance (bottom)

Standard error of the difference between the means 2-sample standard error Each sample has own population mean and error When = n’s, add standard errors of the means together 2 2 % S S _ . . 1, 22 Sy =q|l— ——* (M,-M,) — n n,G n, Equation assumes equal sample size When n, # n,, need to calculate pooled variance Let SPSS do it! ‘ (Ml_Mz) Calculations ‘=, (my—m,) Calculate bottom: Calculate t: Calculate top: (M, -M,) > S(a,-My) =\/ f— (Ml _Mz) S(ml_mZ) Calculate df = (n, — 1)+(n,-1) Look up critical t* Does t value exceed critical value?

Ans: X p(x) X(px) X-u (x-u)? ((x-u)?) (p(x)) 19 0.2 3.8 -1.8 3.24 0.648 20 0.2 4 -0.8 0.64 0.128 21 0.3 6.3 0.2 0.04 0.012 22 0.2 4.4 1.2 1.44 0.288 23 0.1 23 2.2 4.84 0.484 20.8 1.56 w = 208 o= 1.56 g = 125 b. If the manager of Suit World wants to make sure that he has enough suits for the next five days, how many should he buy to stock the store? Ans: 115 suits 3 The Bank of America VP feels that each savings account customer has, on average, three credit cards. The following distribution represents the number of credit cards people own. X 0 1 " 3 4 P(X) 0.18 044 0.27 0.08 0.03 a. Find the mean, variance, and standard deviation. X p(x) x(px) X-u (x-u)* ((x-u)?) (p(x)) 0 0.18 0 -1.34 | 1.7956 0.323208 1 0.44 0.44 034 | 0.1156 0.050864 2 0.27 0.54 0.66 0.4356 0.117612

Q Q= 3 0.08 0.24 1.66 2.7556 0.220448 4 0.03 0.12 2.66 7.0756 0.212268 1.34 0.9244 = 134 = 0.924 = 0.96 4. Flower World determines the probabilities for the number of flower arrangements they deliver each day. X 6 7 8 9 10 P(X) 02 02 03 02 01 a. Find the mean, variance, and standard deviation. X p(x) X(px) X-u (xu)? | ((x-u)?) (p(x)) 6 0.2 1.2 -1.8 3.24 0.648 7 0.2 14 -0.8 0.64 0.128 8 0.3 2.4 0.2 0.04 0.012 9 0.2 1.8 e 1.44 0.288 10 0.1 1 2.2 4.84 0.484 7.8 1.56 n=1=78 o= 1586 o= 125

Question 1 A Telecom service provider claims that individual customers pay on an average 400 rs. per month with standard deviation of 25 rs. A random sample of 50 customers bills during a given month is taken with a mean of 250 and standard deviation of 15. What to say with respect to the claim made by the service provider? Solution: First thing first, Note down what is given in the question: Hy (Null Hypothesis) : p = 400 H; (Alternate Hypothesis): p # 400 (Not equal means either p > 400 or p < 400 Hence it will be validated with two tailed test ) o = 25 (Population Standard Deviation) LoS (x) = 5% (Take 5% if not given in question) n = 50 (Sample size) xbar x~ = 250 (Sample mean) s = 15 (sample Standard deviation) n > = 30 hence will go with z-test Step 1: Calculate z using z-test formula as below: z = (x~ = n)/ (o/Vn) z = (250 - 400) / (25/V50) z = -42 .42 Step 2: get z critical value from z table for a = 5% z critical values = (-1.96, +1.96) to accept the claim (significantly), calculated z should be in between -1,96 < 2 € +1.96 but calculated z (-42.42) < -1.96 which mean reject the null hypothesis

. 2 Region of Rejection Region of Rejection Acceptance Region A +1.96 Two Tailed Test x = 5% z-test example 1 Question 2 From the data available, it is observed that 400 out of 850 customers purchased the groceries online. Can we say that most of the customers are moving towards online shopping even for groceries? Solution: Note down what is given: 400 out of 850 which indicates that this is a proportion problem. Proportion p (small p) = 400/850 = 0.47 Ho (Null Hypothesis): P (capital P) > 0.5 (claim is that most of the customers are moving towards online shopping even for groceries which mean at least 50% should do online shopping) H; (Alternate Hypothesis): P < = 0.5 left tailed n = 850 LoS (o) = 5% (assume 5% as it not given in question) n > = 30 hence will go with z-test

For one or two proportion proportion we use x2 that Team A (Sample A): proportion pa (small p) = na = 1000 Team B (Sample B): proportion pg (small p) = ng = 800 Take a = 5% (assume o = 5% type problem we use z-test. is chi square test) 250/1000 = 0.25 300/800 = 0.375 if not given in question) (in case of multi- Claim: Team B's performance is superior than Team A which means: Hp (Null Hypothesis): Team A (with respect to population) H; (Alternate Hypothesis) Step 1: B > = Bpa (one right tailed test) calculate z value from two proportion z-test formula as below: T — where p* (p hat) = p* (p hat) = 7 = {0.25 = ©Q.315) [/ z = -0.125/[0.212*0.00135] z = -436.75 Step 2: get z using z-table for « Now calculated z -436.75 < +1.645 (1000*0.25 + 80020.375) (ba = pe)/Ip2(1-p*) (1/na + 1/ng)] (na*pa + npg*ps) / (na + npg) (1000 + 800) = [0.305% (1-0.305)*(1/1000 + '1/800) ] 02305 overall mean error of Team B pg < pa overall mean error of = 5% which is z = +1.645 z value fall inside the acceptance region, hence accept the null hypothesis -436.75 Acceptance Region Region 2f Rejection one Tailed Test <= 5% Hence will conclude that null hypothesis is true which mean from given data it is proven significantly that team B's performance is better that team A's performance.