Tutors Answer Your Questions about Hypothesis-testing (FREE)
Question 925703: A local television station wants to estimate the average number of hours people spend watching television each day. They randomly sample 400 individuals in their viewing area and find that, on average, people watch 3.5 hours of television a day. It is known, however, that the population standard deviation is 2.0 hours.
a - determine the standard error of the mean.
b - compute a 99% confidence interval for the mean number of hours spent watching television.
c - interpret the result of the confidence interval in a sentence or two
Click here to see answer by ewatrrr(24785)   |
Question 927434: A manufacturer contemplating the purchase of new tool making equipment has specified that on the average the equipment should not require more than 10 minutes of set up time per hour of operation. The purchasing agent visits a company where the equipment being considered is installed; from records there he notes that 40 randomly selected hours of operation included a total of 7 hours and 30 minutes of set up time, and the standard deviation of set up time per hour was 3 minutes. Based on this sample result, can the assumption that the equipment meets the set up time specification be rejected at the 1 percent level of significance?
Click here to see answer by ewatrrr(24785) |
Question 927878: The average household income per year is μ = $ 15,000. For a random sample of 15 households in the area, the mean x-bar = 14,000 with a sample standard deviation of s = $2,000. Test the hypothesized value of the mean at the 5 percent level of significance.
Click here to see answer by ewatrrr(24785) |
Question 927869: The average household income per year is μ = $ 15,000. For a random sample of 15 households in the area, the mean x-bar = 14,000 with a sample standard deviation of s = $2,000. Test the hypothesized value of the mean at the 5 percent level of significance.
Click here to see answer by ewatrrr(24785) |
Question 939584: A manufacturer of electronic kits has found that the mean time required to assemble circuit tester is 3 hours. with a standard deviation of 0.20 hours. in the test of effectiveness of the new booklet, 15 circuit tester require a mean of 2.90 hours to complete the job. Assuming the population of times is normally distributed, and using the 0.05 level of significance.
a) should we conclude that the new booklet is effective ?
b) known or unknown ?
with solution thank you
Click here to see answer by oscargut(2103)  |
Question 950651: I have a question with answer, But I cant not figure how the number come from.
The number that I don't know is -5 (in the line with *****)
I think it is a silly question, but i really cant figure it out. Thanks
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A sample of 7 students that drank coffee before a test had a mean test mark of 70 with a standard deviation of 10 and a sample of 5 students that did not drink the coffee before the test had a mean of 75 with a standard deviation of 14
For the following calculations You may use the fact that the Standard error of the difference between means is 6.885
Question:
Find the 95% confidence interval for the difference in the group means
Answer:
Use two sided t table value with df =smallest sample size –1= 5-1=4 , column 0.05
This tells the 95% of a sample differences are within 2.776 standard errors of the population difference
So the confidence interval is
between
-5-2.776× 6.885 and -5+2.776×6.885 ***This Line***
Do between –24.1 and 14.1
Click here to see answer by Theo(13342)  |
Question 967336: 4. The IRS claims that it takes 87 minutes to prepare a tax form. You believe it takes longer than 87 minutes. You ask 19 tax filers and find the following preparation times:
61, 74, 78, 79, 87, 88, 96, 97, 98, 98, 99, 100, 100, 101, 108, 110, 117, 120, 121
Determine if the preparation time is significantly longer than 87 minutes. Use α = .025.
Click here to see answer by jim_thompson5910(35256) |
Question 977155:
For this fictional situation, choose whether you should reject or fail to reject the null hypothesis.
Null hypothesis: Making people think about their mortality will not affect their self-esteem.
Experimental outcome: People made to think about their mortality reported lower self-esteem than those that had not thought about their mortality.
Based on the above, we should(1 pt.)
a) fail to reject the null hypothesis
b) reject the null hypothesis
Click here to see answer by solver91311(24713)  |
Question 999697: P&G wants to find out the mean spending on bath soaps by an family per month.P&G management assumed that the average expenditure on bath soaps per month was 90$ and the population standard deviation is 10.Take a sample of 50 families in your town and comment on the acceptance of the survey.
Click here to see answer by ikleyn(52781)  |
Question 1039187: A researcher finds that of 1,000 people who said they attend a religious service at least once a week, 31 stopped to help a person with car trouble. Of 1,200 people interviewed who had not attended a religious service at least once a month, 22 stopped to help a person with car trouble. At the 0.05 significance level, test the claim that the two proportions are equal.
HO:
H1:
Test statistic:
Critical Value:
Click here to see answer by Boreal(15235)  |
Question 1040123: Travel association claims that the mean daily meal cost for two adults traveling together on vacation is $105. A random sample of 15 such groups of adults has a mean daily meal cost of $115 and a standard deviation of $6.50. Is there enough evidence to reject the claim at a=.10?
I know the t value = 5.96 but how do I find the P-VALUE?
Click here to see answer by stanbon(75887) |
Question 1040365: "Using your data from 50 tosses, test the claim that the population proportion of Kisses chocolates that land completely on the base IS 50% at a=10% level of significance." (We tossed the candies and the total # is 13 kisses that landed on the base)
C. Determine standardized test statistic
D. Find P-value
Click here to see answer by stanbon(75887) |
Question 1040525: This is a 2 parter using the same table. I do not understand the concept, can you explain this process and the solution. Thanks
Suppose that for a certain illness, the probability is 55% that a given patient will improve without treatment. The probability that at least n out of 30 patients will improve without treatment is given in the following table.
N 19 20 21 22 23 24 24
Probability that at least n improve 0.23 0.14 0.069 0.031 0.012 0.0040
We give an experimental drug to 30 patients who have this illness. Suppose we count the test significant if the p-value is 0.06 or less. How many patients must show improvement in order to make the test statistically significant?
19.
21.
23.
22.
10. Suppose that for a certain illness, the probability is 55% that a given patient will improve without treatment. The probability that at least n out of 30 patients will improve without treatment is given in the following table.
An experimental drug is given to 30 patients who have this illness. If 20 people show improvement, what is the p-value?
0.23.
0.069.
0.012.
Click here to see answer by rothauserc(4718)  |
Question 1040329: "Using your data from 50 tosses, test the claim that the population proportion of Kisses chocolates that land completely on the base IS 50% at a=10% level of significance." (We tossed the candies and the total # is 13 kisses that landed on the base)
A: State null and alternate hypothesis, identify the claim
B: State level of significance
C. Determine standardized test statistic
D. Find P-value
E. Do you reject Ho or fail to reject Ho
F. Interpret decision in the context of original claim
Click here to see answer by robertb(5830)  |
Question 1041477: Out of this date, whats the range of values that are within 1 standard deviation of the mean. Round to the nearest tenth.
(338, 160, 210, 180, 166, 272, 128, 409, 240, 250, 160, 300, 340, 200, 194, 132, 202, 180, 420, 380, 325, 185, 233, 198)
A.) 241.8 - 328.4
B.) 155.2 - 241.8
C.) 155.2 - 328.4
D.) 328.4 - 415
Click here to see answer by jim_thompson5910(35256) |
Question 1044443: In a study of 420,009 cell phone users, 157 subjects developed brain cancer. Test the claim that cell phone users develop brain cancer at a rate that is different from the rate of 0.0340% for people who do not use cell phones. Because this issue has such great importance, use a 0.005 significance level. Use this information to answer the following questions.
B. What is the test statistic?
C. What is the P-vaule
Click here to see answer by Boreal(15235) |
Question 1047236: 75% of high school students have seen at least one episode of news. Last month, a random sample of 120 high school students were selected and asked if they had seen at least one episode of news. Out of the 120, 85 of them said they had seen at least one episode. Is there enough evidence to claim the population proportion of all high school student that have watched at least one episode is less than 75% at the 0.05 level of significance?
a. Are the requirements met to run a test like this?
b. What are the hypotheses for this test?
c. Compute the test statistic and the p-value for this test.
d. What is your conclusion at the 0.05 level of significance
Click here to see answer by ewatrrr(24785) |
Question 1056584: Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization
thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they
spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test,
the Type I error is:
a. to conclude that the current mean hours per week is higher than 4.5, when in fact, it is higher
b. to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same
c. to conclude that the mean hours per week currently is 4.5, when in fact, it is higher
d. to conclude that the mean hours per week currently is no higher than 4.5, when in fact, it is not higher
Click here to see answer by Boreal(15235) |
Question 1056589: A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She
takes four new students. She records their 18-hole scores before learning the technique and then after having taken her class.
She conducts a hypothesis test. The data are as follows.
Player 1 Player 2 Player 3 Player 4
Mean score before class 83 78 93 87
Mean score after class 80 80 86 86
Table 10.31
The correct decision is:
a. Reject H0.
b. Do not reject the H0.
Click here to see answer by Boreal(15235) |
Question 1056588: A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. Of interest is whether the liquid
diet yields a higher mean weight loss than the powder diet. The powder diet group had a mean weight loss of 42 pounds
with a standard deviation of 12 pounds. The liquid diet group had a mean weight loss of 45 pounds with a standard deviation
of 14 pounds.
Click here to see answer by stanbon(75887) |
Question 1057876: My younger brother had a run in earlier with Médecins Sans Frontières. He narrowly escaped from an adverse verdict by the court........ What he wants is that he be left alone to run his small café......
He asked my oldest brother if he can conduct a survey for him about justice in the Canadian Court.
An initial survey was performed right after Médecins Sans Frontières accused my brother of wrong doing. Of 1852 customers, 53 were against the aggressive tactics of Médecins Sans Frontières. After my brother was cleared by the court, a follow-up survey was performed. Of 4699 customers, 1751 said they did not agree with the aggressive tactics of Médecins Sans Frontières.
My question below please show work
At the 1% significance level, do the data suggest that a higher percentage of customers were against Médecins Sans Frontières after the court case?
Click here to see answer by stanbon(75887) |
Question 1057888: Do men and women select different breakfasts? The breakfasts ordered by randomly selected men and women at a popular breakfast place is shown in Table 11.55. Conduct a test for homogeneity at a 5% level of significance.
French Toast Pancakes Waffles Omelettes
Men 47 35 28 53
Women 65 59 55 60
Click here to see answer by ewatrrr(24785) |
Question 1057880: A geologist collects hand-specimen sized pieces of limestone from a particular area. A qualitative assessment of both texture and color is made with the following results. Is there evidence of association between color and texture for these limestones? Explain your answer.
chart below
color
TEXTURE LIGHT MEDIUM DARK
FINE 4 20 8
MEDIUM 5 23 12
COARSE 21 23 4
Click here to see answer by ewatrrr(24785) |
Question 1058050: Testing the variance
Suppose an airline claims that its flights are consistently
on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no
more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next
25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes.
My questions below
what is the df=_______ ?
Let a= 0.05
Decision ________
Conclusion write out in complete sentence
Click here to see answer by rothauserc(4718) |
Question 1058561: Suppose a group is interested in determining whether
teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses.
Northeast South West Central East
16.3 16.9 16.4 16.2 17.1
16.1 16.5 16.5 16.6 17.2
16.4 16.4 16.6 16.5 16.6
16.5 16.2 16.1 16.4 16.8
x ¯
= ________ ________ ________ ________ ________
s2 = ________ ________ ________ ________ ________
my question below
what is the F statistic =______
Click here to see answer by rothauserc(4718) |
Question 1085768: It is hypothesized that the market share of a corporation should vary more in an industry with active price competition than in one with duopoly and tacit collusion. Suppose that in a study of the steam turbine generator industry, it was found that in 4 years of active price competition, the variance of General Electric's market share was 85.7. In the following 7 years, in which there was duopoly and tacit collusion, this variance was 23.91. Assume that the data can be regarded as an independent random sample from two normal distributions. Test the null hypothesis that the two population variances are equal against the alternative that the variance of market share is higher in years of active price competition. Answer the following, rounding off your answers to two decimal places.
I calculated test statistic to be 85.7/23.91 = 3.5843
And with a 5 % significance level, critical value to be 4.76
I need help finding the P-value for the test
Click here to see answer by rothauserc(4718) |
Question 1091477: In the past, it has taken an express mail company an average of 2 days to deliver packages. After hiring a consulting firm, they want to know if delivery has improved. Let µ = average number of days needed to deliver a package after hiring a consulting firm. The company wants to test the following:
A sample of 25 packages yields xbar = 1.6 and s = 1.5. For α = 0.05, what would you conclude? Assume the relevant population follows a normal random variable.
Click here to see answer by Boreal(15235) |
Question 1092350: Let the specified dimension and standard deviation for a process be 330 and .5 respectively. Four samples A, B, C and D are selected from a lot, measured dimension for these samples are 101.5, 101, 99 and 98 respectively. Which sample is to be rejected
Click here to see answer by Fombitz(32388)  |
Question 1098603: "Dalmatian Darnation," by Kathy Sparling
A greedy dog breeder named Spreckles Bred puppies with numerous freckles
The Dalmatians he sought Possessed spot upon spot The more spots, he thought, the more shekels. His competitors did not agree That freckles would increase the fee. They said, “Spots are quite nice But they don't affect price; One should breed for improved pedigree.” The breeders decided to prove This strategy was a wrong move. Breeding only for spots
Would wreak havoc, they thought. His theory they want to disprove.
They proposed a contest to Spreckles Comparing dog prices to freckles.
In records they looked up One hundred one pups: Dalmatians that fetched the most shekels. They asked Mr. Spreckles to name An average spot count he'd claim To bring in big bucks. Said Spreckles, “Well, shucks, It's for one hundred one that I aim.” Said an amateur statistician Who wanted to help with this mission. “Twenty-one for the sample Standard deviation's ample:
They examined one hundred and one Dalmatians that fetched a good sum.
They counted each spot, Mark, freckle and dot And tallied up every one.
Instead of one hundred one spots They averaged ninety six dots
Can they muzzle Spreckles’ Obsession with freckles Based on all the dog data they've got?
I need help to figure out the step by step solution.
Click here to see answer by richwmiller(17219)  |
Question 1098601: Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Now, what do you think?
I can't for the life of me figure this out step by step
Click here to see answer by MathTherapy(10552)  |
Question 1105407: Judge the following statements as true or false (justify your answer).
a) When the value of the statistic observed in the sample is in the critical region, the null hypothesis must be accepted.
b) The larger the sample size, the smaller the variance of X (x-bar)
c) When performing a test for the mean with known variance the chi-square distribution must be used.
d) When the p-value of the statistic observed in the sample is greater than the level of significance (α) adopted, one must accept Ho.
Click here to see answer by Boreal(15235) |
Question 1105510: The distribution of income (in minimum salaries) of the certain city population is given below:
(%): 54 | 23 | 10 | 5 | 8 |
Income range: (0-2] (2-4] (4-6] (6-10] (No statement]
Get a new income distribution, considering only the people who declare some income. From the distribution obtained by you, calculate:
a) The average income;
b) The proportion of people receiving up to 1.5 minimum wages.
c) The income range corresponding to the richest 10%.
Click here to see answer by Alan3354(69443)  |
Question 1105990: To profitably produce a planned upgrade of a software product you make, you must charge customers $100. Are
your customers willing to pay this much? You contact a random sample of 50 customers and find that 17 would pay
$100 for the upgrade. Find a 95% confidence interval for the proportion of all of your customers (the population)
who would be willing to buy the upgrade for $100.
Click here to see answer by Boreal(15235) |
Question 1110153: A researcher interested in whether people are able to identify emotions correctly in other people when they are extremely tired. It is known that, using a particular method of measurement, the accuracy rating of people in the general population (who are not extremely tired) are normally distributed with a mean of 82 and a variance of 20. In the present study, however, the researcher arranges to test 50 people who had no sleep the previous night. The mean accuracy for these 50 individuals was 78. Using a two-tailed test and the .05 significance level, what should the researcher conclude? (a) Carry out a Z test using the five steps of hypothesis testing. (b)Make a drawing of the distribution involved. (c) Explain your answer to someone who knows about hypothesis testing with a sample of a single individual but who knows nothing about hypothesis testing with a sample of more than one individual. (d) ADVANCED TOPIC: Figure the 95% confidence interval and explain your answer to someone who is familiar with the general logic of hypothesis testing, the normal curve, Z scores, probability, and the idea of a distribution of means, but who has not heard of confidence interval
Click here to see answer by rothauserc(4718) |
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