Decision Reject H0 p value 000 the null is rejected greater than 16 npi5 1 n1

Decision reject h0 p value 000 the null is rejected

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Decision Reject H0 p-value 0.00 the null is rejected. greater than 16. n*pi>5 1 n*(1-pi)>5 1 0.975 adj. for ND -1.96 Right Tailed test CV 1.64 Decision Do not reject H0 p value 0.155 the null is not rejected on is not less than 0.63.
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it follows a normal decision 0.95 adj. for ND -1.64 Right-tailed Test CV 1.28 Decision Do not reject H0 p-value 0.50 , the null is not rejected. not different from 10 lbs. 9 7.6 7.3 1.2 6 8.8 6.8 8.4 6.6 it follows a normal decision 0.75 adj. for ND
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-0.67 Right-tailed Test CV 0.00 Decision Reject H0 p-value 0.02 e null is not rejected. than 6.6. lb. Women Men Label 25 30 mean 14 11 SD 30 25 Sample size est) Women 25 mean 14 std. dev. 30 n 5.000 difference (Men - Women) 3.372 standard error of difference 0 hypothesized difference 1.48 z .1382 p-value (two-tailed) a (0.1), the null is not rejected. n is greater than pop. Of men.
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rtions p2 0.66 0.6657 p (as decimal) 66/100 233/350 p (as fraction) 66 233 X 100 350 n 0.008 difference 0 hypothesized difference 0.0558 std. error 0.14 z .8860 p-value (two-tailed) .5), the null is not rejected. the pop proportion of adults and children. it follows a normal decision 0.99 adj. for ND -2.33 Right-tailed Test CV 2.05 Decision Do not reject H0 p-value 0.42 p c
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p-value Err:502 null is not rejected not less than 60,000. it follows a normal decision 0.99 adj. for ND -2.33 Right-tailed Test CV 2.05 Decision Do not reject H0 p-value 0.04 p-value Err:502 s not rejected not less than 60,000.
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Mean SD
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Mean SD
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Assignment 20 // Pg 366 #'s 46, 48, 50, 52 Question # 46 T-Test One Sample test of hypothesis pop mean, pop SD unknown 321 Step 1 H0: μ ≤ 267 286 290 330 Given Data 310 ypothesized (H0) pop mean mu 267 250 sample mean 288.31 270 sample size 13 280 sample SD 22.4588 299 265 sig level alpha 0.01 291 275 degrees of freedom 12 281 Calculated Test-Statistic T score 3.42 Two tailed test CV 3.0545 -3.0545 Decision Reject H0 p-value 0.0051 Left Tailed Test Right Tailed Test CV -2.68 CV 2.68 Decision DonotrejectH0 Decision Reject H0 p-value 0.0025 p-value 0.0025 Step 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 a H1: μ 267
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T-Test One Sample test of hypothesis 288.30769 mean 22.45879 SD Step 1 Given Data Hypothesized (H0) pop mean mu sample mean sample size sample SD sig level alpha degrees of freedom Calculated Test-Statistic T score Two tailed test CV Decision p-value Left Tailed Test CV Decision p-value Step 5: and there is sufficient evidence to support it.
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Question # 48 pop mean, pop SD unknown H0: μ = 27,000 H1: μ ≠ 27,000 60000 57000 79 12500 0.02 78 -2.13316666015574 2.38 -2.38 DonotrejectH0 0.036055240972422 Right Tailed Test -2.0887 CV 2.0887 Reject H0 Decision DonotrejectH0 0.0180 p-value 0.0180 Since 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.
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H0: μ = 0.63 H1: μ ≠ 0.63 Question # 50 Hypothesis Testing, One sample Proportion Step 1: H0: μ = 0.63 H1: μ ≠ 0.63 Given Data Hypothesized pop. Proportion pi 60,000 sample proportion 57,000 sample size 79 12,500 sig level alpha 0.02 Calculated test statistic z score Err:502 Two Tailed Test CV 2.33 Decision Err:502 p-value Err:502 Left-tailed Test CV -2.05 Decision Err:502 p value Err:502 Step 5: Since test stat (2.51) is greater than CV( We can conclude that the population p
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n*pi>5 1 n*(1-pi)>5 0 0.99 adj. for ND -2.33 Right Tailed test CV 2.05 Decision Err:502 p value Err:502 V(1.96) the null is rejected proportion is not less than 0.63.
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