CH 2/3
Forty-five (45) students at Fullerton College were asked, “What’s your favorite pizza topping?” The results are summarized in the frequency distribution.
a) Fill in the table to the left with the relative frequencies.
b) Is this categorical or quantitative data?
c) What is the mode?
a) check doc
b) Categorical
c) Mode: Pepperoni (most frequent)
According to government data, the probability that an adult was never in a museum is 15%. In a random survey of 10 adults, what is the probability that two or fever were never in a museum?
0.820
From a random sample of 40 watermelons, the mean weight is found to be 22 pounds with a standard deviation of 5.1 pounds.
a) Find a 95% confidence interval for the mean weight of watermelons. Round to one decimal place.
b) Interpret your confidence interval.
c) What is the margin of error for this survey?
a) (20.4,23.6)
b) We are 95% confident that the true mean weight of all the watermelons is between 20.4 and 23.6 pounds
c) 1.6 lbs
The annual salary of employees in a company.
The zip codes of different cities in a state.
The shoe sizes of customers in a shoe store.
Quantitative (Salary is measured in numerical values)
Categorical (Zip codes are numerical but do not represent a measurable quantity)
Quantitative (Shoe sizes follow a numerical scale)
Approximately 65% of U.S. households own a pet.
If three households are selected at random, what is the probability that none of them own a pet?
0.043
A language school tracks the number of new vocabulary words learned by students in their first month. This number is normally distributed with a mean of 95 words and a standard deviation of 15 words.
a) What percent of students learn more than 120 words? (Round to 4 decimal places)
b) The school promises students will learn at least 60 words in their first month. If they don’t, they get a partial refund. If 2,500 students enroll, how many are expected to qualify for a refund?
a)0.0478= 4.78%
b) 57 students
A poll was conducted to determine the proportion of college students who own a laptop. Of the 180 college students surveyed, 68% of them owned a laptop.
a) Construct and interpret a 99% confidence interval for the proportion of college students who own a laptop. Round to three decimal places.
b) If the researchers want the margin of error to be no more than 3% (0.03), what sample size is required, using a prior estimate of 68%?
a) (0.588, 0.768)
b) 1,605
The number of minutes spent studying math per night by a sample of high school students is recorded below.
18, 20, 22, 24, 24, 26, 26, 27, 28, 28, 30, 31, 33, 34, 36, 38, 40, 41
a) Is this categorical or quantitative data?
b) Find the median.
c)Find the mode.
a) Quantitative
b) Median = 28
c) Mode = 24, 26, 28 (all appear twice)
A conference is organizing a lunch menu. There are 12 dessert options, and the organizers will choose 4 different desserts to offer attendees.
495 ways
According to CITA, 72% of adult Americans would rather give up chocolate than their cell phone. In a simple random sample of 300 households, determine the mean and standard deviation number who would rather give up chocolate rather than their cell phone.
216
The scores on a standardized English proficiency test have a mean of 75 and a standard deviation of 12. Suppose a random sample of 81 test scores is taken.
a) Describe the sampling distribution of the sample mean (xˉ) (give the values of μxˉ and σxˉ).
b) What is the probability that the sample mean score is less than 72? Round to 4 decimal places.
c) Would it be unusual for a sample mean score to be greater than 78? Why or why not?
a) sample mean: 75; standard dev. smaple mean= 1.333
b) 0.0122
c) yes, it would be unusual
A health researcher wants to investigate whether there is an association between diet type and self-reported energy levels. A random sample of individuals is surveyed, and the results are shown below:
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a) State the null and alternative hypotheses.
b) What function(s) or steps will you need on the TI-84?
c) Give the P-value. Round to 3 decimal places.
d) Do you reject H₀? Use α = 0.05.
e) State the conclusion.
a) Hypotheses
Null Hypothesis (H₀): There is no association between diet type and self-reported energy level. (They are independent.)
Alternative Hypothesis (H₁): There is an association between diet type and self-reported energy level. (They are dependent.)
b) CHi square independence
c)
c) P-value
Using the steps above, the TI-84 should return something like:χ² test statistic ≈ 3.128
degrees of freedom (df) = (3-1)(3-1) = 4
P-value ≈ 0.537
d) do not reject the null hypothesis.
e) There is not enough evidence at the 0.05 level of significance to conclude that there is an association between diet type and self-reported energy level.
So, we conclude that diet and energy level are independent in this sample.
The favorite color of each student in a class.
The height of basketball players in inches.
The brands of smartphones owned by a group of people.
The number of pets each person in a neighborhood owns.
Categorical (Colors are descriptive labels, not numerical values)
Quantitative (Height is measured numerically)
Categorical (Brands are names, not numbers)
Quantitative (Number of pets is a numerical count)
A wellness coach collected the following data on average nightly sleep hours and stress levels (measured on a 0–10 scale, where 10 is extremely stressed) for six working adults:
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a) Find the linear correlation coefficient between sleep hours and stress level.
(Round to 3 decimal places)
b) Does a linear correlation exist between sleep hours and stress level? Why or why not?
c) Find the equation of the least-squares regression line.
(Round the slope and y-intercept to 3 decimal places)
d) According to the model, what is the predicted stress level for someone who sleeps 6.5 hours per night?
(Round to 2 decimal places)
e) If a person has a stress level of 5.8, how many hours of sleep does the model predict they get?
(Round to 2 decimal places)
a) r≈−0.996
b) ✅ Yes.
The correlation coefficient of –0.996 suggests a very strong negative linear relationship: as sleep hours increase, stress levels tend to decrease.
c) Least-Squares Regression Line
Regression equation:
Stress=−1.118⋅(Sleep Hours)+14.159
✅ Slope ≈ –1.118
✅ Intercept ≈ 14.159
d) Predicted Stress Level for 6.5 Hours of Sleep
Stress=−1.118(6.5)+14.159≈6.89
e) Predicted Sleep Hours for Stress Level of 5.8
Sleep=(5.8−14.159)/1.118≈7.48
Suppose that EverGo Battery Co. produces a rechargeable battery whose life is normally distributed with a mean of 500 hours and a standard deviation of 60 hours.Answer each question using the information above.Write out the function or steps you did on the TI-84, and then circle your final answer.
a) What proportion of EverGo batteries last more than 580 hours? (Round to 4 decimal places.)
b) A battery is considered defective if it lasts less than 410 hours.If EverGo manufactures 25,000 batteries, how many would you expect to be defective?Round to the nearest whole number.
c) A battery is covered under warranty if it lasts fewer hours than a certain threshold. EverGo wants to limit warranty claims to 1% of its batteries.What should they advertise as the warranty life?Round to the nearest 10 hours.
a) 0.0912
b) 1,670 defective batteries
c) 380 hours
A national survey reports that 48% of Americans own a pet dog. A local community group believes that the proportion of dog owners in their town is higher than the national figure. They survey 500 randomly selected residents, and 260 of them own a dog.Is there sufficient evidence to support your suspicion at the 𝛼 = 0.05 level of significance?
a) State the null and alternative hypotheses
b) What is the P-value? (Round to three decimal places)
c) Do you reject H0? Why or why not?
d) State the conclusion
a) H0: p=0.48
H1: p>0,48
b) 0.037
c) Reject H0
d) There is sufficient evidence to support that the proportion of the do owners in the town is greater than the national porportion of 48%
A technology analyst believes that the percentage of college students who own a tablet is higher than the percentage of high school students who own one. To investigate this claim, the analyst collects survey data:
A random sample of 950 college students shows that 418 of them own a tablet.
A random sample of 870 high school students shows that 355 of them own a tablet.
Test the analyst’s claim at the α=0.05 level of significance.
a) Check the requirements for testing the difference between two population proportions.
b) State the null and alternative hypotheses.
c) Calculate the test statistic and the P-value (round to 3 decimal places). Do you reject H0?
d) State your conclusion in the context of the problem.
a) Successes and failures are both ≥ 10 for each group
b)
H0:p1≤p2 (The proportion of college students with tablets is less than or equal to high school)
Hap1>p2 (The proportion of college students with tablets is greater than high school)
c) z≈1.528
p-value≈0.063
d) Do not reject H0H_0H0
Fifty-two (52) Fullerton College students were asked, “What’s your favorite dessert topping?” The results are summarized in the frequency distribution.
a) Fill in the table to the left with the relative frequencies.
b) Is this categorical or quantitative data?
c) What is the mode?
a) check doc
b) Categorical
c) Mode: Chocolate Syrup
A health researcher collected the following data on daily screen time (in hours) and average nightly sleep duration (in hours) from six teenagers:
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a) Find the linear correlation coefficient between screen time and sleep duration.
(Round to 3 decimal places)
b) Does a linear correlation exist between screen time and sleep duration? Why or why not?
c) Find the equation of the least-squares regression line.
(Round the slope and y-intercept to 3 decimal places)
d) According to the model, what is the predicted sleep duration for a teen with 5.5 hours of screen time per day?
(Round to 2 decimal places)
e) If a teen sleeps 6.8 hours per night, how many hours of screen time does the model predict?
(Round to 2 decimal places)
a) r≈−0.998
b) ✅ Yes.
The correlation coefficient of –0.998 indicates a very strong negative linear relationship: as screen time increases, sleep duration tends to decrease.
c) Regression equation:
Sleep=−0.637⋅(Screen Time)+10.054
✅ Slope ≈ –0.637
✅ Intercept ≈ 10.054
d) Predicted Sleep for 5.5 Hours of Screen Time
Sleep=−0.637(5.5)+10.054≈6.55 hours
e) Predicted Screen Time for 6.8 Hours of Sleep
Screen Time=6.8−10.054/0.637≈5.11 hours
The number of people with blood type O-neg based on a simple random sample of size 10 is recorded. According to the American red cross, 7% of the people in the united States have blood type O-negative. Find the probabilities:
a) Exactly 8 people with blood type O-negative.
b) At most 3 people with blood type O-negative.
c) More than 7 people with a blood type of O-negative.
d)Less than 6 people with blood type O-negative.
e)From 4 to 6, inclusively , with blood type O-negative
a) 2.244e-8
b)0.996
c) 2.282e-8
d) 0.99998
e)0.0036
A certain brand of bottled iced tea is supposed to contain 20 ounces of tea. To avoid underfilling penalties, the target mean amount is 20.1 ounces. However, the filling machine introduces some variability.
The quality-control manager wants to verify that the machine is correctly calibrated and that the mean amount is still 20.1 ounces. She randomly samples 18 bottles of iced tea and records the following amounts (in ounces):
19.9, 20.2, 20.0, 20.1, 20.3, 19.8, 20.0, 20.2, 20.0,
20.4, 19.9, 20.3, 20.1, 20.1, 19.7, 20.2, 20.1, 19.9
A normal probability plot suggests the data could come from a normally distributed population
a) State the null and alternative hypotheses
b) What is the P-value? (Round to three decimal places)
c) Does the machine need to be recalibrated? Use α=0.01and explain your answer.
d) State your conclusion
a)
Null hypothesis H0H_0H0: μ=20.1
Alternative hypothesis : μ≠20.1
(This is a two-tailed test.
b) P-value ≈ 0.270
c)we do not reject the null hypothesis.
d) There is not sufficient evidence at the 0.01 significance level to conclude that the mean amount of tea differs from 20.1 ounces.
So, the machine does not need to be recalibrated based on this sample.
In 2019, Congress had the following religious affiliations: 55% Protestant, 31% Catholic, 2% Mormon, 6% Jewish, 2% Muslim, 4% Unaffiliated.
A survey of the public had the following counts: 616 Protestant, 287 Catholic, 20 Mormon, 20 Jewish, 57 Muslim, 193 Unaffiliated.
a) Fill in the expected values.
b) State the null and alternative hypotheses.
c) What function(s) or steps will you need on the TI-84?
d) Give the P-value. Round to 3 decimal places.
e) Do you reject H₀? Use α = 0.01.
f) State the conclusion.
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