Question

In: Statistics and Probability

Question 3 A quality control engineer at a potato chip company tests the bag filling machine...

Question 3

A quality control engineer at a potato chip company tests the bag filling machine by weighing bags of potato chips. Not every bag contains exactly the same weight. But if more than 14% of bags are over-filled then they stop production to fix the machine. They define over-filled to be more than 1 ounce above the weight on the package.

The engineer weighs 175 bags and finds that 56 of them are over-filled. He plans to test the hypotheses H0: p = 0.14 versus Ha: p > 0.14. What is the test statistic?

z =

(Please round your answer to two decimal places if necessary.)

Question 4

A quality control engineer at a potato chip company tests the bag filling machine by weighing bags of potato chips. Not every bag contains exactly the same weight. But if more than 15% of bags are over-filled then they stop production to fix the machine. They define over-filled to be more than 1 ounce above the weight on the package.

The engineer weighs 100 bags and finds that 21 of them are over-filled. He plans to test the hypotheses H0:p=0.15 versus Ha:p>0.15.

What is the test statistic?

  1. z = 1.68
  2. z = -1.68
  3. z = 4
  4. z = -1.47

Question 5

Does secondhand smoke increase the risk of a low weight birth? A baby is “low birth weight” if it weighs less than 5.5 pounds at birth. According to the National Center of Health Statistics, about 7.8% of all babies born in the U.S. are categorized as low birth weight. Researchers randomly select 1200 babies whose mothers had extensive exposure to secondhand smoke during pregnancy. 10.4% of the sample are categorized as low birth weight.

Which of the following are the appropriate null and alternative hypotheses for this research question.

  1. H0: p = 0.078; Ha: p ≠ 0.078
  2. H0: p = 0.078; Ha: p > 0.078
  3. H0: p = 0.104; Ha: p ≠ 0.104
  4. H0: μ = 0.104; Ha: μ > 0.104

Question 6

Short-term classes: Does taking a class in a short-term format (8 weeks instead of 16 weeks) increase a student’s likelihood of passing the course? For a particular course, the pass rate for the 16-week format is 59%. A team of faculty examine student data from 40 randomly selected accelerated classes and determine that the pass rate is 78%.

Which of the following are the appropriate null and alternative hypotheses for this research question?

  1. H0: p = 0.59; Ha: p ≠ 0.59
  2. H0: p = 0.59; Ha: p > 0.59
  3. H0: p = 0.78; Ha: p ≠ 0.78
  4. H0: p = 0.78; Ha: p > 0.59

Question 7

One population proportion test: Which of the following situations involves testing a claim about a single population proportion?

  1. A recent study estimated that 20% of all college students in the United States smoke. The head of Health Services at Goodheart University suspects that the proportion of smokers may be lower there.
  2. A certain prescription allergy medicine is supposed to contain an average of 245 parts per million (ppm) of a certain chemical. The manufacturer takes a sample to check whether the mean concentration is the required 245 ppm or not.
  3. According to the College Board, which administers the SAT exam, the average score for females is lower than for males among graduating seniors in 2011. An educational researcher wants to test whether this is also true in her school district.
  4. A Statistics student at Tacoma Community Colleges wants to determine whether there is a difference in the proportions of class meetings that male and female students attend.

Question 8

According to a Pew Research Center study, in May 2011, 34% of all American adults had a smart phone (one which the user can use to read email and surf the Internet). A communications professor at a university believes this percentage is higher among community college students. She selects 377 community college students at random and finds that 148 of them have a smart phone. Then in testing the hypotheses:

H0: p = 0.34 versus

Ha: p > 0.34,

what is the test statistic?

z =  . (Please round your answer to two decimal places.)

Question 9

According to a Pew Research Center study, in May 2011, 33% of all American adults had a smart phone (one which the user can use to read email and surf the Internet). A communications professor at a university believes this percentage is higher among community college students. She selects 349 community college students at random and finds that 138 of them have a smart phone. In testing the hypotheses:

H0: p = 0.33 versus

Ha: p > 0.33,

she calculates the test statistic as z = 2.5990.

Then the p‑value = .

(Please round your answer to four decimal places.)

Solutions

Expert Solution

Q3: n = 175, x = 56

p̄ = x/n = 0.32

Null and Alternative hypothesis:  

Ho : p = 0.14 ; H1 : p > 0.14

Test statistic:  

z = (p̄ -p)/√(p*(1-p)/n) = (0.32 - 0.14)/√(0.14 * 0.86/175) = 6.86

-------------------------------------------------

Q4: n = 100, x = 21

p̄ = x/n = 0.21

Null and Alternative hypothesis:  

Ho : p = 0.15 ; H1 : p > 0.15

Test statistic:  

z = (p̄ -p)/√(p*(1-p)/n) = (0.21 - 0.15)/√(0.15 * 0.85/100) = 1.68

-------------------------------------------------

Q5: Null and Alternative hypothesis:  

Ho : p = 0.078

H1 : p > 0.078

-------------------------------------------------

Q6: Null and Alternative hypothesis:  

Ho : p = 0.59

H1 : p > 0.59

-------------------------------------------------

Q7: Answer: A recent study estimated that 20% of all college students in the United States smoke. The head of Health Services at Goodheart University suspects that the proportion of smokers may be lower there.

-------------------------------------------------

Q8: n = 377, x = 148

p̄ = x/n = 0.3926

Null and Alternative hypothesis:  

Ho : p = 0.34 ; H1 : p > 0.34

Test statistic:  

z = (p̄ -p)/√(p*(1-p)/n) = (0.3926 - 0.34)/√(0.34 * 0.66/377) = 2.15

-------------------------------------------------

Q9: n = 349, x = 138

p̄ = x/n = 0.3954

Null and Alternative hypothesis:  

Ho : p = 0.33 ; H1 : p > 0.33

Test statistic:  

z = (p̄ -p)/√(p*(1-p)/n) = (0.3954 - 0.33)/√(0.33 * 0.67/349) = 2.5990

p-value = 1- NORM.S.DIST(2.599, 1) = 0.0047


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