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notsponge [240]
3 years ago
12

The power takeoff driveline on tractors used in agriculture is a potentially serious hazard to operators of farm equipment. The

driveline is covered by a shield in new tractors, but for a variety of reasons, the shield is often missing on older tractors. Two types of shields are the bolt-on and the flip-up. It was believed that the bolt-on shield was perceived as a nuisance by the operators and deliberately removed, but the flip-up shield is easily lifted for inspection and maintenance and may be left in place. In a study initiated by the U.S. National Safety Council, random samples of older tractors with both types of shields were taken to see what proportion of shields were removed. Of 183 tractors designed to have bolt-on shields, 35 had been removed. Of the 136 tractors with flip-up shields, 15 were removed. We wish to perform a test of H0: pb = pf versus Ha: pb pf where pb and pf are the proportions of all tractors with the bolt- on and flip-up shields removed, respectively. Which of the following conditions for performing the appropriate significance test is definitely not satisfied in this case?
(a) Both populations are Normally distributed.
(b) The data come from twp independent samples
(c) Both samples were chosen at random.
(d) The counts of successes and failures are large enough to use Normal calculations.
(e) Both populations are at least 10 times the corresponding sample sizes
Mathematics
1 answer:
Roman55 [17]3 years ago
5 0

Answer:

(a) Both populations are Normally distributed.

Step-by-step explanation:

Hello!

There are two types of shields in older tractors, bolt-on shield, and flip-up shield. The researcher wants to study the proportion of older tractors whose shields were removed. For these two random samples where taken, one of older tractors with bolt-on shields and the other one of older tractors with flip-up shields, the number of tractors whose shields were removed was recorded in both groups.

Sample 1: bolt-on

nb= 183

xb= 35 with removed shields

pb'= 35/183= 0.19

Sample 2: flip-up

nf= 136

xf= 15 with removed shields

pf'= 15/136= 0.11

Statistic hypotheses:

H₀: pb=pf

H₁: pb≠pf

(a) Both populations are Normally distributed. d) The counts of successes and failures are large enough to use Normal calculations.

The populations are not normal, these are two binomial populations.

To be able to use the normal calculations the following conditions should be met:

Population 1

n₁≥30 ⇒ nb= 183

n₁*p₁'≥5 ⇒ nb*pb'= 183*0.19= 34.77

n₁*(1-p₁')≥5 ⇒ nb*(1-pb')= 183*(1-0.19)= 148.23

Population 2

n₂≥30 ⇒ nf= 136

n₂*p₂'≥5 ⇒ nf*pf'= 136*0.11= 14.96

n₂*(1-p₂')≥5 ⇒ nf*(1-pf')= 136*(1-0.11)= 121.04

(b) The data come from two independent samples (c) Both samples were chosen at random.

Both conditions are met, the samples are randomly selected, the randomization included in the sampling method guarantees the independence between the samples.

(e) Both populations are at least 10 times the corresponding sample sizes

Without the information regarding the populations, you'll have to assume that the population of old tractors with bolt-on shields is greater than 1830 and the population of old tractors with flip-up shields is greater than in 1360.

I hope this helps!

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ladessa [460]
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3 years ago
"A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour. An approximation is 20 percen
Marizza181 [45]

Answer:

The company should take a sample of 148 boxes.

Step-by-step explanation:

Hello!

The cable TV company whats to know what sample size to take to estimate the proportion/percentage of cable boxes in use during an evening hour.

They estimated a "pilot" proportion of p'=0.20

And using a 90% confidence level the CI should have a margin of error of 2% (0.02).

The CI for the population proportion is made using an approximation of the standard normal distribution, and its structure is "point estimation" ± "margin of error"

[p' ± Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }]

Where

p' is the sample proportion/point estimator of the population proportion

Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} } is the margin of error (d) of the confidence interval.

Z_{1-\alpha /2} = Z_{1-0.05} = Z_{0.95}= 1.648

So

d= Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }

d *Z_{1-\alpha /2}= \sqrt{\frac{p'(1-p')}{n} }

(d*Z_{1-\alpha /2})^2= \frac{p'(1-p')}{n}

n*(d*Z_{1-\alpha /2})^2= p'(1-p')

n= \frac{p'(1-p')}{(d*Z_{1-\alpha /2})^2}

n= \frac{0.2(1-0.2)}{(0.02*1.648)^2}

n= 147.28 ≅ 148 boxes.

I hope it helps!

3 0
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Answer:

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