i) To conclude that Flight time has changed when in fact they didn't
ii) |Z(calc)| = 1.74 < |Z(crit)| = 1.96 => Ho accepted
iii) Accepted Ho => Type II Error

i) Ho: p = 1/8 ; Ha: p> 1/8
ii) 0.1195
iii) 0.1195

i) |Z(calc)| = 1.68 < |Z(crit)| = 1.96 => Ho accepted
ii) Significance level = 0.05

i) Watch the video
ii) 0.0556

0.1172 or 0.12;

i) p-value = 0.1028 > alpha = 0.05 Accept Ho
ii) significance level = 0.05

i) alpha max=0.0434

i) watch the video
ii) 0.0367

i) Not representative sample
ii) alpha=0.02, mean is no longer = 6, when it's not true
iii) p-value = 0.0098 < alpha = 0.02 Reject Ho

i) Ho: h=0.1 ; Ha: h>0.1
ii) p-value = 0.09 < alpha =0.10, Reject Ho

i) mean = 296 ; Var = 187.755 (188)
ii) alpha = 0.85
iii) 0.849

p-value = 0.075 > alpha =0.05, Accept Ho

a) (6.77; 7.43)
b) 0.04^3
c) Not representative, chosen from particular day and time

i) (0.453; 0.577)
ii) 1-0.92=0.08

i) mean = 0.0335 ; Var = 0.0000339
ii) |t(crit)|=1.66 > |t(obs)| = 1.21; Can't reject Ho

i) p-value = 0.08 > alpha = 0.05, Accept Ho
ii) Watch the video
iii) 0.122

i) Watch the video
ii) 0.02^2

Times are independent, p-value = 0.071 > alpha = 0.05, Can't Reject Ho

i) Ho: p=0.3 ; Ha: p <0.3
ii) 0.04023
iii) 0.0823

i) 0.0788
ii) 7489.33 (7490)

i) Probability could be different later during the day or another day
ii) Looking for improvement => Ha: p<0.2(one tailed)
iii) Probability has decreased when in fact it hasn't
iv) X<=2 critical region = 0.04418 = Type I error
v) 3 is not in critical region => Can't Reject Ho

|t(crit)| = 1.664 > |t(obs)|= 1.54, Accept Ho

a) mean = 3.79, s = 0.742
b) |t(crit)| = 1.337 > |t(obs)| = 1.06, Accept Ho

a) p-value = 0.0217 < alpha = 0.05, Reject Ho
b) SRS and independent samples

a) mean= 75.3 ; s = 83.8
b) (72.3 ; 76.62)
c) SRS , n>30

p-value = 0.0262 < alpha = 0.05, Reject Ho

a) Mean(A)= 4.52 ; s^2(A) = 1.52
b) |t(crit)| = 1.671 < |t(obs)| = 1.87 Reject Ho
c) Watch the video
d) 2 independent random samples

(127.2; 151.0)

a) Watch the video
b) Watch the video
c) |t(crit)| = 1.671 < |t(obs)| = 1.89, Reject Ho

Chi Square (crit) = 4.61 > Chi Square (obs) = 3.27, then Accept Ho

Chi(obs) = 14.64 > Chi(crit) = 9.49, Reject Ho

Ho: variables are independent
Ha: they are dependent somehow
Chi(obs) = 5.11 < Chi(crit) = 5.99, Fail to Reject Ho

0.05< P value< 0.1 = alpha, Reject Ho

|t(obs)|= 2.00 < |t(crit)|= 2.262, Accept Ho (at 2.5% alpha)

a) 0.965
b) Lower crit.value = 0.3447 ; Upper = 4.74

a) Watch the formula in the video
b) 0.0818
c) 0.5665

a) Mean upper = 252.4; Mean lower = 247.6
b) 0.0606

a) (3.55; 4.47)
b) 0.30 < sigma ^2 < 2.13
c) 0.0418

a) Mean = 124 , S^2 = 43
b) Chi(obs) = 13.76 < Chi(crit) = 15.51, Accept Ho
c) 4.10 < sigma^2 < 23.6
d) Gurdip's claim is incorrect

a) F(obs) < 6.7572, Acceptance region (accept Ho)
b) d.f = 13, alpha = 5% ; |t(obs)| = 0.79 < |t(crit)| = 1.771, Accept Ho
c) sigma1^2 = sigma2^2 => S(pooled)…

i) Watch the video
ii) Chi(obs) = 6.90 > Chi(crit) = 5.99, Reject Ho

i) |t(obs)| = 2.46 < |t(crit)| = 2.042, Reject Ho (depends on your alpha)
ii) Ha: Mean 1 > Mean 2; |t(obs)| > |t(crit)| = 1.697, Reject Ho (depends on your alpha)

i) Assumption : SRS's from independent populations;
p-value=0.0132 > alpha = 0.01 Accept Ho (depends on your alpha)
ii) (-0.215 ; -0.0073)

i) Watch the video
ii) Chi(obs) = 16.4 > Chi(crit) = 7.81, Reject Ho

i) |t(obs)| = 2.31 > |t(crit)| = 1.86 at 5% sign.lvl, Reject Ho
ii) we assume approx. normal distribution for differences
iii) ( -18.4 ; 14.4)

i) t(obs) = 2.08; at 1% sign.lvl t(crit) = 2.492 Can't Reject ; at 5% sign.lvl t(crit) = 1.711 Reject Ho
ii) Two independent samples have apprxm. normal populations
iii) (0.039 ; 7.82)

Chi (obs) = 1.26 < Chi(crit) = 5.99. The distribution is binomial.

d.f. = 2 (not 3). Chi(obs) = 21.9 > Chi(crit) = 5.99.
It is a strong evidence that the distribution is not Poisson.

a) Ho: Gender and Preferred Subject independent
Ha: they are dependent
b) (130*68)/300
c) Chi(obs) = 8.69
d) Chi(obs) < Chi(crit) = 9.21, Can't Reject Ho
e) Chi(obs) > Chi(crit) = 5.99, Reject Ho

Ho: Die Balanced
Ha: Die unbalanced
Chi(obs) = 13.44 < Chi(crit) =15.09, Can't Reject Ho

i) Ho: Sigma^2 = 5.1
Ha: Sigma^2 < 5.1 (left tail)
Chi(obs) = 25.88 < Chi(crit) = 26.51, Reject Ho
ii) 2.80^2 < sigma^2 < 4.05^2

Ho: Sigma1^2 = Sigma2^2
Ha: Sigma1^2 > Sigma2^2
0.025< P value < 0.05 = alpha Reject Ho

1b) negative;
1c) yes;
2a) watch the video;

watch the video, it is a mistake in the task;

it is better to watch the video!

1b) i) y = -1.28x + 219; ii) x = -0.693y + 160;
2b) r^2 = 1;

Correct Answer: C

Correct Answer: C

Correct Answer: B

Correct Answer: C

Correct Answer: C

Correct Answer: C

Correct Answer: B

Correct Answer: B

Correct Answer: E

Correct Answer: C

Correct Answer: A

Correct Answer: A

Correct Answer: A

Correct Answer: A

Correct Answer: A

Correct Answer: A

Correct Answer: A

Correct Answer: A

Correct Answer: A