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nasty-shy [4]
3 years ago
9

Job-sharing is an innovative employment alternative that originated in Sweden and is becoming popular in the United States (for

example, in 2001, 26% of companies in the U.S. offered it as a flexible work option). Firms that offer job-sharing plans allow two or more persons to work part-time, sharing one full-time job. For example, two job-sharers might alternate work weeks, with one working while the other is off. Job-sharers never work at the same time, and may not even know each other. Job-sharing is particularly attractive to working mothers and to people who frequently lose their jobs due to fluctuations in the economy. In a recent survey of 1,035 major U.S. firms, approximately 23% offer job-sharing to their employees.
For each of the questions, select the correct answer from choices i. through xi. listed below.

a. What is the population of interest?
b. What is the variable measured?
c. What is the sample selected?
d. What is the parameter?
e. What is the statistic?


i. All major U.S. firms
ii. All major firms in the world
iii. All U.S. firms
iv. Job-sharing availability
v. Proportion of employees who are working mothers
vi. The 1,035 firms surveyed
vii. The percentage of all U.S. major firms which o↵er job-sharing to their employees
viii. The percentage of the 1035 major firms surveyed which o↵er job-sharing to their employees
ix. 25%
x. The percentage of all U.S. major firms which employ working mothers
xi. All workers in the U.S. who have participated in he job-sharing program
Mathematics
1 answer:
Veronika [31]3 years ago
3 0

Answer:

a. The population of interest is:

xi. All workers in the U.S. who have participated in the job-sharing program

b. The variable that is being measured is:

viii. The percentage of the 1035 major firms surveyed which offer job-sharing to their employees

c. The sample selected is:

vi. The 1,035 firms surveyed

d. The parameter is:

v. Proportion of employees who are working mothers

e. The statistic is:

ix. 25% (stated as 23%)

Step-by-step explanation:

The parameter of a population is a fixed value calculated from every individual in the population.  It is different from a statistic, which is computed from a sample of the population.  The statistic is meant to approximate a population parameter.

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Answer: We need how much he spent on the whole thing

Step-by-step explanation:

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A student was asked to find a 90% confidence interval for the proportion of students who take notes using data from a random sam
Scorpion4ik [409]

Answer:

After use the formula we got the following result for the 90% confidence interval (0.13 <p<0.34)

And the conclusion for this case would be:

e. With 90% confidence, the proportion of all students who take notes is between 0.13 and 0.34.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Description in words of the parameter p

p represent the real population proportion of students who take notes

\hat p represent the estimated proportion of students who take notes

n is the sample size required  

z_{\alpha/2} represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Numerical estimate for p

In order to estimate a proportion we use this formula:

\hat p =\frac{X}{n} where X represent the number of people with a characteristic and n the total sample size selected.

Confidence interval

The confidence interval for a proportion is given by this formula  

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}  

For the 90% confidence interval the value of \alpha=1-0.90=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.64  

After use the formula we got the following result for the 90% confidence interval (0.13 <p<0.34)

And the conclusion for this case would be:

e. With 90% confidence, the proportion of all students who take notes is between 0.13 and 0.34.

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Bobby drove 110 miles, and his car used up 5 gallons of gas. How many miles can he drive with 16 gallons of gas?
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110/5 = 22

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Answer:

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