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polet [3.4K]
3 years ago
15

Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the

night shift than on the day shift. A sample of 60 day-shift workers showed that the mean number of units produced was 334, with a population standard deviation of 23. A sample of 68 night-shift workers showed that the mean number of units produced was 341, with a population standard deviation of 28 units.At the .10 significance level, is the number of units produced on the night shift larger?1. This is a (Click to select)twoone-tailed test.2. The decision rule is to reject H0: μd ≥ μn if z < . (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)3. The test statistic is z = . (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)4. What is your decision regarding H0?
Mathematics
1 answer:
Alborosie3 years ago
8 0

Answer:

Step-by-step explanation:

Let the subscripts d and n represent day and night respectively

The null hypothesis is

H0 : μd ≥ μn

The alternative hypothesis is

H1 : μd < μn

it is a one-tailed and also a right left test because of the greater than symbol in the alternative hypothesis.

The decision rule is to reject H0: μd ≥ μn If 0.10 > p value

Since the population standard deviations are known, we would use the formula to determine the test statistic(z score)

z = (xd - xn)/√σd²/nd + σn²/nn

Where

xd and xn represents sample means for day and night respectively.

σd and σn represents population standard deviations for day and night respectively.

nd and nn represents number of samples

From the information given,

xd = 334

xn = 341

σd = 23

σ2 = 28

nd = 60

nn = 68

z = (334 - 341)/√23²/60 + 28²/68

= - 7/√20.34607843138

z = - 1.55

From the normal distribution table, the probability value corresponding to the z score is 0.061

Since the level of significance, 0.1 > 0.061, we would reject H0

Therefore, there is enough evidence to conclude that there are more units produced on the night shift than on the day shift.

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6 0
3 years ago
It is estimated that 75% of all young adults between the ages of 18-35 do not have a landline in their homes and only use a cell
Mademuasel [1]

Answer:

a) 75

b) 4.33

c) 0.75

d) 3.2 \times 10^{-13} probability that no one in a simple random sample of 100 young adults owns a landline

e) 6.2 \times 10^{-61} probability that everyone in a simple random sample of 100 young adults owns a landline.

f) Binomial, with n = 100, p = 0.75

g) 4.5 \times 10^{-8} probability that exactly half the young adults in a simple random sample of 100 do not own a landline.

Step-by-step explanation:

For each young adult, there are only two possible outcomes. Either they do not own a landline, or they do. The probability of an young adult not having a landline is independent of any other adult, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

75% of all young adults between the ages of 18-35 do not have a landline in their homes and only use a cell phone at home.

This means that p = 0.75

(a) On average, how many young adults do not own a landline in a random sample of 100?

Sample of 100, so n = 100

E(X) = np = 100(0.75) = 75

(b) What is the standard deviation of probability of young adults who do not own a landline in a simple random sample of 100?

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100(0.75)(0.25)} = 4.33

(c) What is the proportion of young adults who do not own a landline?

The estimation, of 75% = 0.75.

(d) What is the probability that no one in a simple random sample of 100 young adults owns a landline?

This is P(X = 100), that is, all do not own. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 100) = C_{100,100}.(0.75)^{100}.(0.25)^{0} = 3.2 \times 10^{-13}

3.2 \times 10^{-13} probability that no one in a simple random sample of 100 young adults owns a landline.

(e) What is the probability that everyone in a simple random sample of 100 young adults owns a landline?

This is P(X = 0), that is, all own. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{100,0}.(0.75)^{0}.(0.25)^{100} = 6.2 \times 10^{-61}

6.2 \times 10^{-61} probability that everyone in a simple random sample of 100 young adults owns a landline.

(f) What is the distribution of the number of young adults in a sample of 100 who do not own a landline?

Binomial, with n = 100, p = 0.75

(g) What is the probability that exactly half the young adults in a simple random sample of 100 do not own a landline?

This is P(X = 50). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 50) = C_{100,50}.(0.75)^{50}.(0.25)^{50} = 4.5 \times 10^{-8}

4.5 \times 10^{-8} probability that exactly half the young adults in a simple random sample of 100 do not own a landline.

8 0
2 years ago
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