A 5

Bouchra Tannir 260356824 Assignment 5 8.32 Cooling method for gas turbines (check which ones is H0) a.

x=11066

s=1594 n=47 α=0.05

H₀: μ≤10000 Ha: μa>10000 Z= x-10000sn= 11066-10000159467=5.47 5.47>1.96=Zα2 Z>Zα2

Thus we reject the null hypothesis H₀ at the significance level α = 0.05. We conclude that the mean heat rate of gas turbines augmented with high pressure inlet fogging exceeds 10000kJ/kWh. b. We rejected H₀ at the significance level α = 0.05. However, we know that this hypothesis testing will lead to this conclusion incorrectly (type I error) only 5% of the time when H₀ is true, meaning when μ is lower than 10000kJ/kWh. If the test statistics does not fall into the rejection region, we do not reject H₀. However, we do not conclude that the null hypothesis is true because we do not know the probability β that our test procedure will lead to an incorrect acceptance of H₀, meaning a type II error. 8.68 Testing a mosquito repellent a. H₀: μ0≥0.95

Ha: µa <0.95

t = x-µ₀ s/√n=0.83-0.950.15/√5= -1.79

- tα = -1.53 -1.79 < -1.53 thus

–t<-tα

α=0.1

Thus we reject H₀ with a significance level α=0.10. We then conclude that the mean repellency of the new mosquito repellent is less than 95% (with a significance level α=0.10). b. The assumptions required are that the random sample is selected from the target population and that the population from which the sample is selected has a distribution that is approximately normal (the sample is smaller than 30). 8.72 Lengths of great White Sharks x=24+20+223=22 S=2 α=0.05 tα2=4.303 H0: μ0≤21 Ha: μa>21 t=x-μ0sn=22-2123=0.866 <4.303= tα2

Thus, we don’t reject H₀. The following data doesn’t support the marine biologist’s claim that the great white shark grows much longer than 21 feet. However, we reserve judgment on which hypothesis is true. 8.86 Verbs and double-object datives H₀: p₀≥ Ha: p<

13 13

σp= 1323√35=0.08

p= 1035=0.29 Z= 0.29-(13)0.08=-0.54 Zα=z0.05=0.0199

-0.54< -0.0199

Thus we reject H₀ with a significance level α=0.05. We then conclude that the true fraction of sentences with the verb “buy” that are DODs is less tha 1/3 (with a significance level α=0.05).

8.152 NCAA March Madness a. 1 vs. 16 seed: H0: p > p₀ HA: p ≤ p₀

p₀ = 0.5

p= 5252

Z = (^p – p₀)/√([ p₀ (1- p₀)/n) Z = (1 – 0.5)/√([0.5 (1- 0.5)/52) ≈ 7.21 Z < -zα Z < -z0.05 7.21 > -1.644 thus H₀ is not rejected. 2 vs. 15 seed: H₀: p > p₀ Ha: p ≤ p₀

p₀ = 0.5

p= 4952

Z = (^p – p₀)/√([ p₀ (1- p₀)/n) Z = (0.9423 – 0.5)/(0.06934) ≈ 6.38 Z < -zα Z < -z0.05 6.38 > -1.644 thus H₀ is not rejected. 8 vs. 9 seed: H0: p > p₀ HA: p ≤ p₀

p₀ = 0.5

p= 2252

Z = (^p – p₀)/√([ p₀ (1- p₀)/n) Z = (0.423 – 0.5)/(0.06934) ≈ -1.11 Z < -zα Z < -z0.05 -1.11 > -1.644 thus H₀ is not rejected. There is not enough evidence to reject the perception that the higher seeded team has a better than 5252 chances of winning. The trend is that as p decreases, the test statistic also decreases. b. 1 vs. 16 seed:

H0: μ > μ₀ HA: μ ≤ μ₀

μ₀ = 10;

x

= 22.9; s = 12.4; n = 52

Z = (x – μ₀)/(s/√n) = (22.9 – 10)/(12.4/√52) ≈ 7.502 Z < -zα Z < -z0.05 7.502 > -1.644 thus H₀ is not rejected. 4 vs. 13 seed: H0: μ > μ₀ HA: μ ≤ μ₀

μ₀ = 10;

x

= 10; s = 12.5; n = 52

Z = (x – μ₀)/(s/√n) = (10 – 10)/(s/√n) =0 Z < -zα Z < -z0.05 0 >-1.644 thus H₀ is not rejected. There is no evidence to reject the claim that a 1-, 2-, 3- or 4seeded team will win by an average of more than 10 points. c. 5 vs. 12 seed: H0: μ < μ₀ HA: μ ≥ μ₀

μ₀ = 5;

x

= 5.3; s = 10.4; n = 52

Z = (x – μ₀)/(s/√n) = (5.3 – 8)/(10.4/√52) ≈ -1.87 Z > zα Z > z0.05 -1.87 < 1.644 thus H₀ is not rejected. 8 vs. 9 seed: H0: μ < μ₀ HA: μ ≥ μ₀ μ₀ = 5; Z = (x – μ₀)/(s/√n)

x

= -2.1; s = 11.0; n = 52

= (-2.1 – 8)/(11/√52) ≈ -6.62 Z > zα Z > z0.05 -6.62 < 1.644 thus H₀ is not rejected. There is not enough evidence to reject claims that 5-, 6-, 7- or 8- seeded teams will win by an average of less than 5 points. e. H₀: μd= μ₀ Ha: μd ≠ μ₀

xd=

0.7; s d = 11.3; n = 360; μ₀ = 0

Z = (xd – μ₀)/(s d /_n) = (0.7 – 0)/(11.3/√360) = (0.7)(√360)/11.3 ≈ 1.175 Z>Zα/2 Z>Z0.05/2

1.175 < 1.96 thus H₀ is not rejected, the true main difference μdcan equal μ₀ = 0. Therefore, the point spread, on average, can be considered a good predictor of the outcome of the game. 9.22 Pig castration study S1=S2=0.9 α=0.05 tα2=t0.025 Spooled= n1-1S12+n2-1S22n1+n2-2=230.92+240.9247=0.9 t=x1-x2- D0Sp21n1+1n2= 0.74-0.70-0√(0.81)(124+125)=0.156 2.00
Thus, we do not reject H₀ and conclude that the population mean number of high frequency vocal responses doesn’t differ for piglet castration by the two methods. 9.48 Homophone Confusion to Alzheimer’s patients. In this case, the two observations are paired (it’s paired difference experiment). The two tests are dependant.

95% confidence interval for

μd= μ2-μ1

xd±tα2Sdn= 1.65±2.0933.2020=0.15, 3.15

This interval doesn’t include 0. Therefore, μ2-μ1>0 and conclude that the patients show an increase in mean homophone confusion errors over time.

μ2>μ1.

We

For this procedure to be valid, we assume that the population of differences has a distribution that is approximately normal and that the sample differences are randomly selected from the population differences (the sample n is small).

The histograms for both samples are not mound shaped, the distribution is skewed, meaning the data is not normally distributed (however we notice that the histogram on the right is more mound shaped). Thus, the assumptions are not satisfied. 9.70 Gambling in Public High Schools. Ho:p1-p2=0 Ha: p1-p2≠0 p1=468421484=0.218 and q1=0.782 p2=531323199=0.23 and q2=0.77 σp1-p2= p1q1n1+p2q2n2=0.2180.78221484+0.230.7723199=0.00395

Z=p1-p2σp1-p2=0.218-0.230.00395=-3.04 -3.04>0.004 Z>Z0.01 thus Z>Z0.005

Thus, we reject the null hypothesis H₀. We conclude that the percentages of ninth grade boys who gambled weekly or daily on any game in 1992 and 1998 are significantly different since p1-p2>0 and p1>p2 and -3.039>0.004 provides strong evidence against it. b.

p-value=2×PZ>-3.04=0.0024

p-value is smaller than professor.

α=0.01.

Therefore, we agree with the

9.114 Hull Failure Oil Tankers Data: Diagram 1 – Spillage by collisions and fires

FIRE: Mean 69.25 Std Dev 61.363485 Std Err Mean 17.714112 Upper 95% Mean 108.2385 Lower 95% Mean 30.261502 N 12

COLLISIONS: Mean 76.6 Std Dev 70.362869 Std Err Mean 22.250693 Upper 95% Mean 126.93456 Lower 95% Mean 26.265436 N 10

HULLFAIL: Mean 58.583333 Std Dev 57.544383 Std Err Mean 16.611633 Upper 95% Mean 95.14529 Lower 95% Mean 22.021376 N 12

GROUNDING: Mean 47.785714 Std Dev 28.466387 Std Err Mean 7.6079618 Upper 95% Mean 64.221717 Lower 95% Mean 31.349712 N 14 a.

Spooled2=0.1161.362+ 970.36220=4298

We consider the variances equal since

S1S2>0.75

x1- x2±tα2Sp21n1+1n2 69.25-76.6±1.7254298112+110 -55.77, 41.07

Since the 90% confidence interval includes 0, we can conclude that there is no evidence of a difference between the mean spillage amount in accidents caused by collision and the mean spillage amount in accidents caused by fires or explosions. b. This case is a small sample with different variances and different samples.

Ho:μ1-μ2=0 Ha: μ1-μ2≠0 T=x1-x2S12n1+S22n2= 58.58-47.7957.54212+28.47214=0.59 ϑ=S12n1+S22n22S12n12n1-1+S22n22(n2-1)=15.523≈16 0.59<2.120 thus 0.59
Therefore, we don’t reject the null hypothesis H₀. We can then conclude that there is no evidence of a difference between the two means and that μ1-μ2=0. c. Assumptions made in part a and part bare the following: the two samples selected from the two populations are independent (samples are also less than 30) and the two populations have normal distributions. For the first diagram, the data is not perfectly symmetrical (the mean is not identical to the median but it is relatively close). As for the second diagram, the mean is equal to the median so we can conclude that the distribution of the data is mound shaped and is in fact a normal distribution. We can assume that the assumptions are reasonably satisfied. 9.134 Students Attitudes towards Parents n=Zα22σD2SE2= 1.6452120.32=30.066≈31

Thus, the number of male required to obtain a 90% confidence interval with a sampling error of 0.3 is 31.