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This homework (among 17 others) was first solved on May 17th, 2022 by our USA essay writers.

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1.  Private nonprofit four-year colleges charge, on average, $26,367 per year in tuition and fees. The standard deviation is $7,114. Assume the distribution is normal. Let X be the cost for a randomly selected college. Round all answers to 4 decimal places where possible.

a. What is the distribution of X? X ~ N(,)

b. Find the probability that a randomly selected Private nonprofit four-year college will cost less than 30,629 per year.

c. Find the 68th percentile for this distribution. $ (Round to the nearest dollar.) 
2. Los Angeles workers have an average commute of 30 minutes. Suppose the LA commute time is normally distributed with a standard deviation of 13 minutes. Let X represent the commute time for a randomly selected LA worker. Round all answers to 4 decimal places where possible.

a. What is the distribution of X? X ~ N(,)

b. Find the probability that a randomly selected LA worker has a commute that is longer than 36 minutes.

c. Find the 85th percentile for the commute time of LA workers.  minutes 
3.  
The average price of a college math textbook is $158 and the standard deviation is $24. Suppose that 49 textbooks are randomly chosen. Round all answers to 4 decimal places where possible.

What is the distribution of ¯xx¯? ¯xx¯ ~ N(,)
For the group of 49, find the probability that the average price is between $157 and $162. 
Find the third quartile for the average textbook price for this sample size. $ (round to the nearest cent)
For part b), is the assumption that the distribution is normal necessary? YesNo

The average price of a college math textbook is $158 and the standard deviation is $24. Suppose that 49 textbooks are randomly chosen. Round all answers to 4 decimal places where possible.

What is the distribution of ¯xx¯? ¯xx¯ ~ N(,)
For the group of 49, find the probability that the average price is between $157 and $162. 
Find the third quartile for the average textbook price for this sample size. $ (round to the nearest cent)
For part b), is the assumption that the distribution is normal necessary? YesNo

4.   
Suppose that the amount of time that students spend studying in the library in one sitting is normally distributed with mean 48 minutes and standard deviation 16 minutes. A researcher observed 14 students who entered the library to study. Round all answers to 4 decimal places where possible.

What is the distribution of XX? XX ~ N(,)
What is the distribution of ¯xx¯? ¯xx¯ ~ N(,)
What is the distribution of ∑x∑x? ∑x∑x ~ N(,)
If one randomly selected student is timed, find the probability that this student’s time will be between 49 and 51 minutes. 
For the 14 students, find the probability that their average time studying is between 49 and 51 minutes. 
Find the probability that the randomly selected 14 students will have a total study time more than 728 minutes. 
For part e) and f), is the assumption of normal necessary? YesNo
The top 15% of the total study time for groups of 14 students will be given a sticker that says “Great dedication”. What is the least total time that a group can study and still receive a sticker? minutes

Suppose that the amount of time that students spend studying in the library in one sitting is normally distributed with mean 48 minutes and standard deviation 16 minutes. A researcher observed 14 students who entered the library to study. Round all answers to 4 decimal places where possible.

What is the distribution of XX? XX ~ N(,)
What is the distribution of ¯xx¯? ¯xx¯ ~ N(,)
What is the distribution of ∑x∑x? ∑x∑x ~ N(,)
If one randomly selected student is timed, find the probability that this student’s time will be between 49 and 51 minutes. 
For the 14 students, find the probability that their average time studying is between 49 and 51 minutes. 
Find the probability that the randomly selected 14 students will have a total study time more than 728 minutes. 
For part e) and f), is the assumption of normal necessary? YesNo
The top 15% of the total study time for groups of 14 students will be given a sticker that says “Great dedication”. What is the least total time that a group can study and still receive a sticker? minutes

5.  
Suppose that the age of students at George Washington Elementary school is uniformly distributed between 6 and 10 years old. 36 randomly selected children from the school are asked their age. Round all answers to 4 decimal places where possible.

What is the distribution of XX? XX ~ U(,)

Suppose that 36 children from the school are surveyed. Then the sampling distribution is

What is the distribution of ¯xx¯? ¯xx¯ ~ N(,)
What is the probability that the average of 36 children will be between 8 and 8.6 years old? 

6.  
A newsgroup is interested in constructing a 95% confidence interval for the proportion of all Americans who are in favor of a new Green initiative. Of the 592 randomly selected Americans surveyed, 370 were in favor of the initiative. Round answers to 4 decimal places where possible.

a.  With 95% confidence the proportion of all Americans who favor the new Green initiative is between  and .
b.  If many groups of 592 randomly selected Americans were surveyed, then a different confidence interval would be produced from each group. About  percent of these confidence intervals will contain the true population proportion of Americans who favor the Green initiative and about  percent will not contain the true population proportion.
7.  A political candidate has asked you to conduct a poll to determine what percentage of people support her. If the candidate only wants a 0.5% margin of error at a 99.5% confidence level, what size of sample is needed? When finding the z-value, round it to four decimal places. 
8.  
A fitness center is interested in finding a 98% confidence interval for the mean number of days per week that Americans who are members of a fitness club go to their fitness center. Records of 269 members were looked at and their mean number of visits per week was 2.5 and the standard deviation was 2.9. Round answers to 3 decimal places where possible.

a. To compute the confidence interval use a ? t z  distribution.
b. With 98% confidence the population mean number of visits per week is between  and   visits.
c. If many groups of 269 randomly selected members are studied, then a different confidence interval would be produced from each group. About  percent of these confidence intervals will contain the true population mean number of visits per week and about  percent will not contain the true population mean number of visits per week. 
9.  
A 2011 survey, by the Bureau of Labor Statistics, reported that 91% of Americans have paid leave. In January 2012, a random survey of 1000 workers showed that 89% had paid leave.
The resulting p-value is 0.0271; thus, the null hypothesis is rejected. It is concluded that there has been a decrease in the proportion of people, who have paid leave from 2011 to January 2012.
What type of error is possible in this situation?

type I
type II
neither
both

10.  
The average wait time to get seated at a popular restaurant in the city on a Friday night is 9 minutes.  Is the mean wait time less for men who wear a tie? Wait times for 12 randomly selected men who were wearing a tie are shown below. Assume that the distribution of the population is normal.
8, 8, 6, 8, 7, 8, 9, 10, 7, 8, 6, 9
What can be concluded at the the αα = 0.05 level of significance level of significance?

For this study, we should use Select an answer t-test for a population mean z-test for a population proportion 
The null and alternative hypotheses would be:     

H0:H0:  ? μ p  Select an answer ≠ > = <     H1:H1:  ? p μ  Select an answer ≠ < > =    

The test statistic ? t z  =  (please show your answer to 3 decimal places.)
The p-value =  (Please show your answer to 4 decimal places.)
The p-value is ? ≤ >  αα
Based on this, we should Select an answer accept reject fail to reject  the null hypothesis.
Thus, the final conclusion is that …

The data suggest the population mean is not significantly less than 9 at αα = 0.05, so there is statistically insignificant evidence to conclude that the population mean wait time for men who wear a tie is equal to 9.
The data suggest the populaton mean is significantly less than 9 at αα = 0.05, so there is statistically significant evidence to conclude that the population mean wait time for men who wear a tie is less than 9.
The data suggest that the population mean wait time for men who wear a tie is not significantly less than 9 at αα = 0.05, so there is statistically insignificant evidence to conclude that the population mean wait time for men who wear a tie is less than 9.

11.  
The average retirement age in America is 64 years old. Do small business owners retire at a different average age? The data below shows the results of a survey of small business owners who have recently retired. Assume that the distribution of the population is normal.
68, 68, 53, 54, 63, 58, 50, 68, 53, 62, 60, 52, 61, 58
What can be concluded at the the αα = 0.10 level of significance level of significance?

For this study, we should use Select an answer z-test for a population proportion t-test for a population mean 
The null and alternative hypotheses would be:     

H0:H0:  ? μ p  Select an answer ≠ < > =    
H1:H1:  ? p μ  Select an answer ≠ = > <     The test statistic ? t z  =  (please show your answer to 3 decimal places.) The p-value =  (Please show your answer to 4 decimal places.) The p-value is ? ≤ >  αα
Based on this, we should Select an answer fail to reject reject accept  the null hypothesis.
Thus, the final conclusion is that …

The data suggest the populaton mean is significantly different from 64 at αα = 0.10, so there is sufficient evidence to conclude that the population mean retirement age for small business owners is different from 64.
The data suggest the population mean is not significantly different from 64 at αα = 0.10, so there is sufficient evidence to conclude that the population mean retirement age for small business owners is equal to 64.
The data suggest the population mean retirement age for small business owners is not significantly different from 64 at αα = 0.10, so there is insufficient evidence to conclude that the population mean retirement age for small business owners is different from 64.

12.  
Only 14% of registered voters voted in the last election. Will voter participation increase for the upcoming election? Of the 362 randomly selected registered voters surveyed, 69 of them will vote in the upcoming election. What can be concluded at the αα = 0.05 level of significance?

For this study, we should use Select an answer t-test for a population mean z-test for a population proportion 
The null and alternative hypotheses would be:     

H0:H0:  ? p μ  Select an answer > ≠ = <   (please enter a decimal)    H1:H1:  ? p μ  Select an answer = < > ≠   (Please enter a decimal)

The test statistic ? z t  =  (please show your answer to 3 decimal places.)
The p-value =  (Please show your answer to 4 decimal places.)
The p-value is ? ≤ >  αα
Based on this, we should Select an answer reject fail to reject accept  the null hypothesis.
Thus, the final conclusion is that …

The data suggest the populaton proportion is significantly higher than 14% at αα = 0.05, so there is statistically significant evidence to conclude that the the percentage of all registered voters who will vote in the upcoming election will be higher than 14%.
The data suggest the population proportion is not significantly higher than 14% at αα = 0.05, so there is statistically insignificant evidence to conclude that the percentage of registered voters who will vote in the upcoming election will be higher than 14%.
The data suggest the population proportion is not significantly higher than 14% at αα = 0.05, so there is statistically significant evidence to conclude that the percentage of registered voters who will vote in the upcoming election will be equal to 14%.

 

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