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alexgriva [62]
3 years ago
13

A research team investigates the potential association between hours of work and smoking. They want to see whether there is a di

fference in number of smokes per day between smokers who work about 40 hours per week and those who work 55 hours or more. If they were to use the hypothesis testing procedure for this research question, what would be the appropriate test statistic to use
Mathematics
1 answer:
Goryan [66]3 years ago
8 0

Answer:

chi-square

Step-by-step explanation:

According to the given situation, Chi-square statistic would be considered for as we have to determine whether working hours and smoking are attached with each other or not.

To know whether working hours based on smoking or there is an important relationship between both- So Chi-square statistic would be used

Therefore the test that should be used is chi-square

You might be interested in
An office supply company ordered 24 cartons of specialty paper. The regular price for a carton of specialty paper is $6.88. They
zimovet [89]

Answer: A. $5 × 25

Step-by-step explanation:

Given the following :

Number of cartons of specialty paper ordered = 24 cartons

Regular price for a carton = $6.88

Discount given off each carton = $1.75

Hence,

Discounted price per carton = $(6.88 - 1.75) = $5.13

Actual amount to be paid for the order :

Discounted price * number of cartons to be ordered

$5.13 * 24

Hence, a reasonable estimate will be :

$5.13 = $5( nearest whole number)

24 can be rounded to 25 (this covers for the $0.13 on the discounted cost per carton)

Reasonable estimate :

$5 × 25

7 0
3 years ago
Read 2 more answers
The function f(x) = –x2 − 2x + 15 is shown on the graph. What are the domain and range of the function?
Makovka662 [10]
The function is a parabola which opens downwards.
-x^2 - 2x + 15
-(x^2 + 2x) + 15
= -[(x + 1)^2 - 1 ]+ 15
= -(x + 1)^2 + 16

The domain is all real numbers and the range is {y|y <= 16}
Its the second choice.


4 0
4 years ago
Please help will give brainliest answer
Natasha_Volkova [10]

Answer:

B

Step-by-step explanation:

oh my goodness I don't know

7 0
3 years ago
The number of accidents per week at a hazardous intersection varies with mean 2.2 and standard deviation 1.4. This distribution
ankoles [38]

Answer:

a) Normal Distribution

b) 0.146

c) 0.070

Step-by-step explanation:

We are given the following information in the question:

The number of accidents per week at a hazardous intersection varies with mean 2.2 and standard deviation 1.4

a) \bar{x}: mean number of accidents per week at the intersection during a year (52 weeks)

According to central limit theorem, as the sample size becomes larger, the distribution of mean approaches a normal distribution.

Since we have a large sample, the approximate distribution of \bar{x} is a normal distribution with

\text{Mean} = 2.2\\\text{Standard Deviation} = \displaystyle\frac{\sigma}{\sqrt{n}} = \frac{1.4}{\sqrt{52}} = 0.19

b) P(mean is less than 2)

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

P(x > 610)

P( \bar{x} < 2) = P( z < \displaystyle\frac{2 - 2.2}{0.19}) = P(z < -1.052)

Calculation the value from standard normal z table, we have,  

P(\bar{x} < 2) = 0.146

c) P(fewer than 100 accidents at the intersection in a year)

P(x < 100)

P( \bar{x} < \frac{100}{52}) =P(\bar{x} < 1.92) =P(z < \displaystyle\frac{1.92 - 2.2}{0.19}) = P(z < -1.473)

Calculation the value from standard normal z table, we have,  

P(x

7 0
3 years ago
Mr.Hall stands 200 meters from the base of the water tower. He finds the angle of elevation to the top of the water to be 27 deg
gayaneshka [121]

Answer:

227

Step-by-step explanation:

Just add :)

5 0
4 years ago
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