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MATH 225N Week 7 Hypothesis Testing Questions
and Answers
Week 7 Hypothesis Testing Q & A
1. Steve listens to his favorite streaming music service when he works out. He wonders whether the service algorithm does a good job of finding random songs that he will like more often than not. To test this, he listens to 50 songs chosen by the service at random and finds that he likes 32 of them. Use Excel to test whether Steve will like a randomly selected song more than not and then draw a conclusion in the context of a problem. Use α = 0.05. Type equation here. Ho: p = ≤ 0. 5 (50%) p = 0. Ha: p = > 0.5 (strictly <>≠ ) P-value = 0.02 which is < 𝛼 = 0. 05 we reject Ho and support the Ha Hypothesis Test for p population proportion Level of Significance 0.05 (decimal) Proportion under H 0 0.5000 (decimal) n 50 Number of Successes 32 Sample Proportion 0. StDev 0. SE 0. Test Statistic (z) 1. One-Sided p-value 0. Two-Sided p-value 0. Right-Tailed (>) 1. Left-Tailed (<) - 1. Two-Tailed (≠) ± 1. Answer: Reject the null hypothesis. There is sufficient evidence to prove that Steve will like a random selected song more often than not. 2. A magazine regularly tested products and gave the reviews to its customers. In one of its reviews, it tested 2 types of batteries and claimed that the batteries from company A outperformed batteries from company B in 108 of the tests. There were 200 tests. Company B decided to sue the magazine, claiming that the results were not significantly different from 50% and that the magazine was slandering its good name. Use Excel to test whether the true proportion of times that Company A’s batteries outperformed Company B’s batteries is different from 0.5. Identify the p=value rounding it to 3 decimal places.
Ho: p = 0.5 Ha ≠ 0. 5 (two tailed test) n = 200 (𝛼 𝑖𝑠 𝑛𝑜𝑡 𝑔𝑖𝑣𝑒𝑛 𝑠𝑜 𝑙𝑒𝑎𝑣𝑒 𝑖𝑡 0. 05 ) Hypothesis Test for p population proportion Level of Significance 0. Proportion under H 0 0. n 200 Number of Successes 108 Sample Proportion 0. StDev 0. SE 0. Test Statistic (z) 1. One-Sided p-value 0. Two-Sided p-value 0. Right-Tailed (>) 1. Left-Tailed (<) - 1. Two-Tailed (≠) ± 1. Answer: 0.258 (because it is a two tailed test). We are not rejecting the null hypothesis and we do not have evidence to support the alternative hypothesis.
3. A candidate in an election lost by 5.8% of the vote. The candidate sued the state and said that more than 5.8% of the ballots were defective and not counted by the voting machine, so a full recount would need to be done. His opponent wanted to ask for the case to be dismissed, so she had a government official from the state randomly select 500 ballots and count how many were defective. The official found 21 defective ballots. Use Excel to test if the candidates claim is true and that < 5.8% of the ballots were defective. Identify the p=value rounding to 3 decimal places. Ho: p = ≥ 0. 058 Ha < 0. 058 (one tailed test) n = 500 (𝛼 𝑖𝑠 𝑛𝑜𝑡 𝑔𝑖𝑣𝑒𝑛 𝑠𝑜 𝑙𝑒𝑎𝑣𝑒 𝑖𝑡 0. 05 ) Hypothesis Test for p population proportion Level of Significance 0.05 (decimal) Proportion under H 0 0.0580 (decimal) n 500 Number of Successes 21 Sample Proportion 0. StDev 0. SE 0. Test Statistic (z)
One-Sided p-value 0. Two-Sided p-value 0.
6. An economist claims that the proportion of people that plan to purchase a fully electric vehicle as their next car is greater than 65%. To test this claim, a random sample of 750 people were asked if they planned to purchase a fully electric vehicle as their next car. Of this 750, 513 indicated that they plan to purchase an electric vehicle. Ho: p = 0.65 Ha; p = >0. Find the p-value for this hypothesis test for a proportion & round to 3 decimal places. Hypothesis Test for p population proportion Level of Significance 0. Proportion under H 0 0. n 750 Number of Successes 513 Sample Proportion 0. StDev 0. SE 0. Test Statistic (z) 1. One-Sided p-value 0. Two-Sided p-value 0. Answer: 0. 7. Colton makes the claim to his classmates that < 50% of newborn babies born this year in his state are boys. To prove this claim, he selects a random sample of 344 birth records in his state from this year. Colton found that 176 of the newborns were boys. What are the null and alternative hypothesis for this hypothesis test. Answer: Ho: 0. Ha: <0. 8. An Airline company claims that in its recent advertisement that at least 94% of passenger luggage that is lost is recovered and reunited with their customer within 1 day. Hunter is a graduate student studying statistics. For a research project, Hunter wants to find out whether there is sufficient evidence in support of the airline company’s claim. He randomly selects 315 passengers whose luggage was lost by the airlines and found out that 276 of those passengers were reunited with their luggage within 1 day. Are all of the conditions for his hypotheses test met, and if so, what are the Ho and Ha for this hypothesis test? For a binomial Model to follow the normal model, the following condition must be satisfied: Success count = n * p ≥ 5 and Failure count ≥ 5 Example: success count 315 * 0.94 = 296.1 and failure count 315-296.1 = 18.9 so it meets the conditions.
Answer: All of the conditions were met and the Ho = 0.94; Ha = >0.
9. A college administrator claims that the proportion of students who are nursing majors is > 40%. To test this claim, a group of 400 students are randomly selected and its determined that 190 are nursing majors. The following is the set up for the hypothesis test: Ho: p = .40 and Ha: p = >. Find the test statistics for this hypothesis test for a proportion & round to 2 decimal places. Answer: 3. Level of Significance 0. Proportion under H 0 0. n 400 Number of Successes 190 Sample Proportion 0. StDev 0. SE 0. Test Statistic (z) 3. One-Sided p-value 0. Two-Sided p-value 0. 10. A hospital administrator claims that the proportion of knee surgeries that are successful are 87%. To test this claim, a random sample of 450 patients who underwent knee surgery is taken and it is determined that 371 patients had a successful knee surgery operation. Ho: p = 0.87 Ha: p ≠ 𝟎. 𝟖𝟕 (two sided tail) Find the test statistics for this hypothesis test for a proportion & round to 2 decimal places. Answer: - 2.87 (this would be rejected) Level of Significance 0. Proportion under H 0 0. n 450 Number of Successes 371 Sample Proportion 0. StDev 0. SE 0. Test Statistic (z) - 2. One-Sided p-value 0. Two-Sided p-value 0.
alternative hypothesis, Ha = 𝝁 = > $𝟓𝟓, 𝟎𝟎𝟎 , find the test statistic 𝝉 and the p-value for the appropriate hypothesis test. Round the 𝝉 to 2 decimal places and the p-value to 3 decimal places. Hypothesis Test for μ for unknown population stdev Level of Significance 0.1 (decimal) Mean under H 0 55000 n 61 Sample Mean 56500 StDev 3750 SE 480. Test statistic (t) 3. One-Sided p-value 0. Two-Sided p-value 0. Answer: t = 3.12; p-value = 0.
- A college administrator claims that the proportion of students that are nursing majors is less than 40%. To test this claim, a group of 400 students are randomly selected and its determined that 149 are nursing majors. The following is the setup for this hypothesis test: H 0: p =0. Ha : p <0. Find the p-value for this hypothesis test and round your answer to 3 decimal places. Hypothesis Test for p population proportion Level of Significance 0. Proportion under H 0 0. n 400 Number of Successes 149 Sample Proportion 0. StDev 0. SE 0. Test Statistic (z)
One-Sided p-value 0. Two-Sided p-value 0.
Answer: 0.
18. A researcher claims that the incidence of a certain type of cancer is less than 5%. To test
this claim, the a random sample of 4000 people are checked and 170 are determined to
have the cancer. The following is the setup for this hypothesis test:
H 0 :p=0.
Ha:p<0.
In this example, the p-value was determined to be 0..
Come to a conclusion and interpret the results for this hypothesis test for a proportion (use a
significance level of 5% )
Select the correct answer below: The decision is to reject the Null Hypothesis. The conclusion is that there is enough evidence to support the claim. (p=0.015 𝛼 = 0. 05 ) The decision is to fail to reject the Null Hypothesis. The conclusion is that there is not enough evidence to support the claim.
( So, if p≤α , reject H 0 ; otherwise fail to reject H 0 )
19. A police office claims that the proportion of people wearing seat belts is less than 65%.
To test this claim, a random sample of 200 drivers is taken and its determined
that 126 people are wearing seat belts.
The following is the setup for this hypothesis test:
H 0 :p=0.
Ha:p<0.
In this example, the p-value was determined to be 0..
Come to a conclusion and interpret the results for this hypothesis test for a proportion (use a
significance level of 5% )
Select the correct answer below: The decision is to reject the Null Hypothesis. The conclusion is that there is enough evidence to support the claim.
The following is the setup for this hypothesis test:
H 0 :p=0.
Ha:p<0.
In this example, the p-value was determined to be 0..
Come to a conclusion and interpret the results for this hypothesis test for a proportion (use a
significance level of 5 )
Select the correct answer below: The decision is to reject the Null Hypothesis. The conclusion is that there is enough evidence to support the claim. The decision is to fail to reject the Null Hypothesis. The conclusion is that there is not enough evidence to support the claim.
- A researcher claims that the proportion of people who are right-handed is 70%. To test this claim, a random sample of 600 people is taken and its determined that 397 people are right handed. The following is the setup for this hypothesis test: H 0: p = 0.
Ha : p ≠ 0.
Find the p-value for this hypothesis test for a proportion and round your answer to 3 decimal places. Answer: 0. Hypothesis Test for p population proportion Level of Significance 0. Proportion under H 0 0. n 600 Number of Successes 397 Sample Proportion 0. StDev 0. SE 0. Test Statistic (z)
One-Sided p-value 0. Two-Sided p-value 0.
24. Kathryn, a golfer, has a sample driving distance mean of 187.3 yards from 13 drives.
Kathryn still claims that her average driving distance is 207 yards, and the low average
can be attributed to chance. At the 1% significance level, does the data provide sufficient
evidence to conclude that Kathryn's mean driving distance is less than 207 yards? Given
the sample data below, accept or reject the hypothesis.
H 0 :μ=207 yards; Ha:μ<207 yards
α=0.01 (significance level)
z 0 =−1.
p=0.
Select the correct answer below: Reject the null hypothesis because the p-value 0.0721 is greater than the significance level α=0.01. Do not reject the null hypothesis because the p-value 0.0721 is greater than the significance level α=0.01. Reject the null hypothesis because |−1.46|>0.01. Do not reject the null hypothesis because |−1.46|>0.01. Do not reject the null hypothesis because the value of z is negative.
25. Mary, a javelin thrower, claims that her average throw is 61 meters. During a practice
session, Mary has a sample throw mean of 55.5 meters based on 12 throws. At
the 1% significance level, does the data provide sufficient evidence to conclude that
Mary's mean throw is less than 61 meters? Accept or reject the hypothesis given the
sample data below.
H 0 :μ=61 meters; Ha:μ<61 meters
α=0.01 (significance level)
z 0 =−1.
p=0.
Select the correct answer below: Reject the null hypothesis because |−1.99|>0.01. Do not reject the null hypothesis because |−1.99|>0.01. Reject the null hypothesis because the p-value 0.0233 is greater than the significance level α=0.01. Do not reject the null hypothesis because the value of z is negative. Do not reject the null hypothesis because the p-value 0.0233 is greater than the significance level α=0.01.
28. Ruby, a bowler, has a sample game score mean of 125.8 from 25 games. Ruby still
claims that her average game score is 140 , and the low average can be attributed to
chance. At the 5% significance level, does the data provide sufficient evidence to
conclude that Ruby's mean game score is less than 140? Given the sample data below,
accept or reject the hypothesis.
H 0 :μ=140 ; Ha:μ<
α=0.05 (significance level)
z 0 =−0.
p=0.
Select the correct answer below: Reject the null hypothesis because the value of z is negative. Do not reject the null hypothesis because |−0.52|>0.05. Reject the null hypothesis because the p-value 0.3015 is greater than the significance level α=0.05. Do not reject the null hypothesis because the p-value 0.3015 is greater than the significance level α=0.05. Reject the null hypothesis because |−0.52|>0.05.
29. Timothy, a bowler, has a sample game score mean of 202.1 from 11 games. Timothy still
claims that his average game score is 182 , and the high average can be attributed to
chance. At the 5% significance level, does the data provide sufficient evidence to
conclude that Timothy's mean game score is greater than 182? Given the sample data
below, accept or reject the hypothesis.
H 0 :μ=182 ; Ha:μ>
α=0.05 (significance level)
z 0 =1.
p=0.
Select the correct answer below: Reject the null hypothesis because the p-value 0.0582 is greater than the significance level α=0.05. Do not reject the null hypothesis because the p-value 0.0582 is greater than the significance level α=0.05. Do not reject the null hypothesis because 1.57>0.05. Reject the null hypothesis because the value of z is positive. Reject the null hypothesis because 1.57>0.05.
30. What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z 0 =−1.59? (Do
not round your answer; compute your answer using a value from the table below.) Table score was 0.056.
Provide your answer below:.
31. What is the p - value of a right-tailed one-mean hypothesis test, with a test statistic of z 0 =2.05? (Do not
round your answer; compute your answer using a value from the table below.) The number on the table was . 1 - .980 =. Answer: 0.
31. What is the p - value of a left-tailed one-mean hypothesis test, with a test statistic
of z 0 =−1.19?
Use a p - value rounded to 3 decimal places. P = 0.
32. What is the p - value of a left-tailed one-mean hypothesis test, with a test statistic of z 0 =−0.65?
(Do not round your answer; compute your answer using a value from the table below. Answer: 0.
33. What is the p - value of a right-tailed one-mean hypothesis test, with a test statistic of z 0 =0.36?
(Do not round your answer; compute your answer using a value from the table below.) Number in table. 1 – 0.641 =. Answer: 0.
34. What is the p - value of a right-tailed one-mean hypothesis test, with a test statistic of z 0 =1.82?
(Do not round your answer; compute your answer using a value from the table below.) Number in table was 0.966. So 1-.966 = 0. Answer: 0.
35. What is the p - value of a two-tailed one-mean hypothesis test, with a test statistic of z 0 =−1.73?
(Do not round your answer; compute your answer using a value from the table below.) Table # was 0.042 so 0.042 * 2 = 0. Answer: 0.
- A CEO wondered if her company received either more or less complaints from its workers
on Monday than any other day. She figured that if it were truly random, 20% of the
complaints should have been filed on Monday. She randomly selected 50 complaints and
38. A CEO wondered if her company received either more or less complaints from its workers
on Monday than any other day. She figured that if it were truly random, 20% of the
complaints should have been filed on Monday. She randomly selected 50 complaints and
checked the day that they were submitted. In those complaints 13 were submitted on a
Monday.
The CEO conducts a one-proportion hypothesis test at the 5% significance level, to test whether the true
proportion of complaints submitted on a Monday is different from 20%.
(a) H 0 :p=0.2 ; Ha:p≠0.2 , which is a two-tailed test.
(b) z 0 =1.061 , p - value is = 0.
(c) Which of the following are appropriate conclusions for this hypothesis test? Select all that apply. Select all that apply:
We reject H 0.
We fail to reject H 0.
At the 5% significance level, the data provide sufficient evidence to conclude the true proportion is
different than 20%.
At the 5% significance level, the data do not provide sufficient evidence to conclude the true proportion is
different than 20%.
39. A business owner claims that the proportion of take out orders is greater than 25%. To
test this claim, the owner checks the next 250 orders and determines that 60 orders
are take out orders. The following is the setup for this hypothesis test:
{H 0 :p=0.25Ha:p>0.
Find the test statistic for this hypothesis test for a proportion. Round your answer
to 2 decimal places.
Hypothesis Test for p population proportion Level of Significance 0. Proportion under H 0 0. n 250 Number of Successes 60 Sample Proportion 0.
StDev 0. SE 0. Test Statistic (z)
One-Sided p-value 0. Two-Sided p-value 0. Provide your answer below: -.
40. Colton makes the claim to his classmates that less than 50% of newborn babies born
this year in his state are boys. To prove this claim, he selects a random sample
of 344 birth records in his state from this year. Colton found that 176 of the newborns
are boys. What are the null and alternative hypotheses for this hypothesis test? Select the correct answer below: H 0 :p≠0.5 Ha:p=0. H 0 :p=0.5 Ha:p≠0. H 0 :p=0.5 Ha:p>0. H 0 :p=0.5 Ha:p<0.
41. Kylie works for a large nursery and is investigating whether to use a new brand of seeds.
The new brand of seeds advertises that 93% of the seeds germinate, which is higher
than the germination rate of the seeds she is currently using. She will change over to this new brand unless the actual germination rate is less than what is advertised. Kylie
conducts an experiment by randomly selecting 76 seeds of the new brand and plants
them. She finds that 70 of those seeds germinated. What are the null and alternative
hypotheses for this hypothesis test? Select the correct answer below: H 0 :p=0.93 Ha:p>0. H 0 :p=0.93 Ha:p<0. H 0 :p≠0.93 Ha:p=0. H 0 :p=0.93 Ha:p≠0.
42. The owners of a supermarket chain are looking into the effectiveness of the supermarket's loyalty card program. Specifically, they would like to know if the percentage of shoppers
in their stores who use the loyalty card has changed from 63% in 2014. Chloe works in
the marketing department of the chain and is assigned to answer the owners' inquiry. She
Sample Proportion 0. StDev 0. SE 0. Test Statistic (z) 1. One-Sided p-value 0. Two-Sided p-value 0. Provide your answer below: 1.
45. Rosetta, a pitcher, claims that her pitch speed is more than 57 miles per hour, on
average. Several of her teammates do not believe her, so she decides to do a hypothesis
test, at a 1% significance level, to persuade them. She throws 10 pitches. The mean
speed of the sample pitches is 64 miles per hour. Rosetta knows from experience that the
standard deviation for her pitch speed is 4 miles per hour.
H 0 : μ≤57; Ha: μ>
α=0.01 (significance level)
What is the test statistic (z-score) of this one-mean hypothesis test, rounded to two decimal places?
Provide your answer below: 5. Hypothesis Test for μ Population stdev known Level of Significance 0. Mean under H 0 57 n 10 Sample Mean 64 StDev 4 SE 1. Test Statistic (z- score) 5. One-Sided p-value 0. Two-Sided p-value 0.
46. Which of the following results in a null hypothesis p≤0.61 and alternative hypothesis p>0.61?
Select the correct answer below: A study says that at least 61% of students study less than 5 hours per week. A researcher thinks this is incorrect, and wants to show that fewer than 61% of students study less than 5 hours per week.
A study says that more than 61% of students study less than 5 hours per week. A researcher thinks this is incorrect, and wants to show that at least 61% of students study less than 5 hours per week. A study says that at most 61% of students study less than 5 hours per week. A researcher thinks this is incorrect, and wants to show that more than 61% of students study less than 5 hours per week. A study says that less than 61% of students study less than 5 hours per week. A researcher thinks this is incorrect, and wants to show that more than 61% of students study less than 5 hours per week.
47. Suppose the null hypothesis, H 0 , is: a weightlifting bar can withstand weights of 800 pounds
and less. What is α , the probability of a Type I error in this scenario?
the probability that you think the weightlifting bar can withstand weights of 800 pounds and less when, in fact, it cannot the probability that you think the weightlifting bar can withstand weights of 800 pounds and less when, in fact, it can the probability that you think the weightlifting bar cannot withstand weights of 800 pounds and less when, in fact, it can the probability that you think the weightlifting bar cannot withstand weights of 800 pounds and less when, in fact, it cannot
48. Suppose a pitcher claims that his pitch speed is less than 43 miles per hour, on average.
Several of his teammates do not believe him, so the pitcher decides to do a hypothesis
test, at a 10% significance level, to persuade them. He throws 19 pitches. The mean
speed of the sample pitches is 35 miles per hour. The pitcher knows from experience that
the standard deviation for his pitch speed is 6 miles per hour.
H 0 : μ≥43; Ha: μ<
α=0.1 (significance level)
What is the test statistic (z-score) of this one-mean hypothesis test, rounded to two decimal places?
Hypothesis Test for μ Population stdev known Level of Significance 0. Mean under H 0 43 n 19 Sample Mean 35 StDev 6 SE 1.
