Question

12. An agricultural research company has developed two new types of soy bean seeds, call them "seed A" and "seed B". A study is conducted to determine which will produce a higher mean yield. To test the two types of seed, 20 similar plots of land were randomly placed into one of two groups. One group of ten plots was planted with "seed A", while the other ten plots were planted with "seed B". The yield of each field, in bushels per acre, was recorded in the table below.

Seed A	101	109	88	108	112	105	119	99	95	112
Seed B	91	102	103	105	87	97	84	88	100	94

1. Conduct a hypothesis test at a 0.050.05 level of significance to determine if the two types of soy beans produce different mean yields.

The test statistic is ________________________
The p-value is____________________________

Construct a 9595% confidence interval for the mean of the differences. Hint: with the data in your lists, use the two-independent sample t-INTERVAL option on your calculator.
__________________to________________________

  

13. A company owns 9 trucks of various makes and models. The manager recently heard that inflating tires with nitrogen may provide slightly better gas mileage. The manager wants to determine if there is a noticeable increase in the mean gas mileage for the 9 trucks when nitrogen is utilized. Over a period of time, a test is run in which the gas mileage of each truck is recorded both with and without nitrogen in the tires. The gas mileages of the 15 trucks with and without nitrogen in the tires are recorded here. (data is in miles per gallon)

Truck	A	B	C	D	E	F	G	H	I
Without Nitrogen	25	20	20	16	25	21	23	24	17
With Nitrogen	28	22	21	19	26	23	23	26	17

(b) The test statistic is_____________________

(c) The p-value is________________________

14. A professor of nursing wonders if the female nursing students are more likely to drop out of a nursing program than the male nursing students. To check her intuition, several nursing programs are compiled and random samples of both male and female nursing students are selected. Of the 200 male nursing students selected, 17 of them did not attain their nursing degree. Of the 700 female nursing students selected, 68 of them did not attain their nursing degree. Test the claim that the proportion of females not completing their degree is higher than the proportion of males using a level of significance of 0.05.
The test statistic is _________________
The p-value is ____________________

Answer 1

12.

For Seed A :

∑x = 1048

∑x² = 110590

n1 = 10

Mean , x̅1 = Ʃx/n = 1048/10 = 104.8000

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(110590-(1048)²/10)/(10-1)] = 9.1869

For Seed B :

∑x = 951

∑x² = 90933

n2 = 10

Mean , x̅2 = Ʃx/n = 951/10 = 95.1000

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(90933-(951)²/10)/(10-1)] = 7.4005

--

Null and Alternative hypothesis:

Ho : µ1 = µ2

H1 : µ1 ≠ µ2

Pooled variance :

S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((10-1)*9.1869² + (10-1)*7.4005²) / (10+10-2) = 69.5833

Test statistic:

t = (x̅1 - x̅2) / √(s²p(1/n1 + 1/n2 ) = (104.8 - 95.1) / √(69.5833*(1/10 + 1/10)) = 2.6002

df = n1+n2-2 = 18

p-value :

Two tailed p-value = T.DIST.2T(ABS(2.6002), 18) = 0.0181

Decision:

p-value < α, Reject the null hypothesis

95% Confidence interval for the difference:

S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((10-1)*9.1869² + (10-1)*7.4005²) / (10+10-2) = 69.583333

Lower Bound = (x̅1 - x̅2) - t-crit*√(S²p*(1/n1 +1/n2)) = (104.8 - 95.1) - 2.101*√(69.5833*(1/10 + 1/10)) = 1.8625

Upper Bound = (x̅1 - x̅2) + t-crit*√(S²p*(1/n1 +1/n2)) = (104.8 - 95.1) + 2.101*√(69.5833*(1/10 + 1/10)) = 17.5375

1.8625 < µ1 - µ2 < 17.5375

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13.

Without Nitrogen	With Nitrogen	Difference
25	28	-3
20	22	-2
20	21	-1
16	19	-3
25	26	-1
21	23	-2
23	23	0
24	26	-2
17	17	0

∑x = -14

∑x² = 32

n = 9

Mean , x̅d = Ʃx/n = -14/9 = -1.5556

Standard deviation, sd = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(32-(-14)²/9)/(9-1)] = 1.1304

Null and Alternative hypothesis:

Ho : µd = 0

H1 : µd < 0

Test statistic:

t = (x̅d)/(sd/√n) = (-1.5556)/(1.1304/√9) = -4.1284

df = n-1 = 8

p-value = T.DIST(-4.1284, 8, 1) = 0.0017

Decision:

p-value < α, Reject the null hypothesis

----------------------

14.

Male nursing student:

n1 = 200, x1 = 17

p̂1 = x1/n1 = 0.085

Female nursing student:

n2 = 700, x2 = 68

p̂2 = x2/n2 = 0.0971

Null and Alternative hypothesis:

Ho : p1 = p2

H1 : p1 < p2

Pooled proportion:

p̄ = (x1+x2)/(n1+n2) = (17+68)/(200+700) = 0.09444

Test statistic:

z = (p̂1 - p̂2)/√ [p̄*(1-p̄)*(1/n1+1/n2)] = (0.085 - 0.0971)/√[0.0944*0.9056*(1/200+1/700)] = -0.5179

z = (p̂1 - p̂2)/√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ] = (0.085 - 0.0971)/√[(0.085*0.915/200) + (0.0971*0.9029/700)] = -0.5355

p-value = NORM.S.DIST(-0.5179, 1) = 0.3023

Decision:

p-value > α, Do not reject the null hypothesis