Decision Reject H0p-value0.00the null is rejected.greater than 16.n*pi>51n*(1-pi)>510.975 adj. for ND-1.96Right Tailed test CV1.64Decision Do not reject H0p value 0.155the null is not rejected on is not less than 0.63.
it follows a normal decision0.95 adj. for ND-1.64Right-tailed TestCV1.28Decision Do not reject H0p-value0.50, the null is not rejected.not different from 10 lbs. 22.214.171.124126.96.36.199.6it follows a normal decision0.75 adj. for ND
-0.67Right-tailed TestCV0.00Decision Reject H0p-value0.02e null is not rejected.than 6.6. lb.WomenMenLabel2530mean1411SD3025Sample sizeest)Women25 mean14 std. dev.30 n5.000 difference (Men - Women)3.372 standard error of difference0 hypothesized difference1.48 z.1382 p-value (two-tailed)a (0.1), the null is not rejected.n is greater than pop. Of men.
rtionsp20.660.6657p (as decimal)66/100233/350p (as fraction)66233X100350n0.008 difference0 hypothesized difference0.0558 std. error0.14 z.8860 p-value (two-tailed).5), the null is not rejected.the pop proportion of adults and children. it follows a normal decision0.99 adj. for ND-2.33Right-tailed TestCV2.05Decision Do not reject H0p-value0.42pc
p-valueErr:502null is not rejected not less than 60,000.it follows a normal decision0.99 adj. for ND-2.33Right-tailed TestCV2.05Decision Do not reject H0p-value0.04p-valueErr:502s not rejected not less than 60,000.
Assignment 20 // Pg 366 #'s 46, 48, 50, 52Question # 46T-Test One Sample test of hypothesis pop mean, pop SD unknown 321Step 1H0: μ ≤ 267286290330Given Data310ypothesized (H0) pop mean mu267250sample mean 288.31270sample size13280sample SD22.4588299265sig level alpha 0.01291275degrees of freedom12281Calculated Test-Statistic T score 3.42Two tailed test CV3.0545-3.0545Decision Reject H0p-value 0.0051Left Tailed Test Right Tailed Test CV-2.68CV2.68Decision DonotrejectH0Decision Reject H0p-value 0.0025p-value0.0025Step 5:since test stat (3.42) is less than CV (2.68), the null is rejected We can conclude that the population mean fare has increased aH1: μ ＞ 267
T-Test One Sample test of hypothesis 288.30769 mean22.45879 SDStep 1Given DataHypothesized (H0) pop mean musample mean sample sizesample SDsig level alpha degrees of freedomCalculated Test-Statistic T score Two tailed test CVDecision p-value Left Tailed Test CVDecision p-value Step 5:and there is sufficient evidence to support it.
Question # 48pop mean, pop SD unknown H0: μ = 27,000H1: μ ≠ 27,000600005700079125000.0278-2.133166660155742.38-2.38DonotrejectH00.036055240972422Right Tailed Test -2.0887CV2.0887Reject H0Decision DonotrejectH00.0180p-value0.0180Since test stat (-2.13) is less than CV(2.38) the null is not rejected We can conclude that the population mean is less than 60,000.
H0: μ = 0.63H1: μ ≠ 0.63Question # 50Hypothesis Testing, One sample ProportionStep 1:H0: μ = 0.63H1: μ ≠ 0.63Given Data Hypothesized pop. Proportion pi60,000sample proportion 57,000 sample size7912,500 sig level alpha 0.02Calculated test statistic z scoreErr:502Two Tailed Test CV2.33Decision Err:502p-valueErr:502Left-tailed TestCV-2.05DecisionErr:502p value Err:502Step 5:Since test stat (2.51) is greater than CV(We can conclude that the population p
n*pi>51n*(1-pi)>500.99 adj. for ND-2.33Right Tailed test CV2.05Decision Err:502p value Err:502V(1.96) the null is rejected proportion is not less than 0.63.