Tutors Answer Your Questions about Hypothesis-testing (FREE)
Question 1192252: In a 1868 paper, German physician Carl Wunderlich reported based on over a million body temperature readings that the mean body temperature for healthy adults is 98.6° F. However, it is now commonly believed that the mean body temperature of a healthy adult is less than what was reported in that paper. To test this hypothesis a researcher measures the following body temperatures from a random sample of healthy adults.
98.2, 98.5, 98.6, 98.5, 97.3, 98.3
(a) Find the value of the test statistic.
Click here to see answer by Boreal(15235)   |
Question 1192845: Ocho Rios Autoworks, a chain of automotive tune-up shops, advertises that its personnel can change oil, replace the oil filter, and lubricate any standard automobile in 15minutes, on the average. The Bureau of Standards received complaints from customers that service
takes considerably longer. To check the firm's claim, the Bureau had service done on 21 unmarked cars. The mean service time was 17 minutes, and the standard deviation of the sample was 1 minute. Use the 0.05 level of significance to check the reasonableness of the
claim made by Ocho Rios Autoworks.
Z = (X - U) / (SD / √n)
=17-15/1/sqrt(21)
= 9.165
I have never gotten a z value so large before, please advise if I am doing something wrong
Click here to see answer by Theo(13342)  |
Question 1193394: 1.The mean number of close friends for the population of people living in the Philippines is 5. The standard deviation of scores in this population is 1.2. An investigator predicts that the mean number of close friends for intorverts will be significantly different from the mean of the population. The mean number of close friends for a sample of 26 introverts is 6. Find the z-value *
Click here to see answer by Boreal(15235) |
Question 1194538: The profits of a corner store for 6 days are 2,000 pesos, 3,000 pesos, 1,200 pesos, 6,000 pesos,
4,000 pesos, and 5,100 pesos. Does this set of data present sufficient evidence that the average
profit per day of the store is more than 3,800 pesos per day? Test at the 0.05 level of
significance.
Click here to see answer by Boreal(15235) |
Question 1194813: Some manufacturers claim that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones. Suppose that consumers test 21 hybrid sedans and get a mean of 31 mpg with a standard deviation of seven mpg. 18non-hybrid sedans get a mean of 22 mpg with a standard deviation of four mpg. At 0.05 level of significance, conduct a hypothesis test to evaluate the manufacturers claim.
Click here to see answer by Boreal(15235) |
Question 1195831: A study was conducted to determine weight loss, body composition, etc. in obese women before and after 12 weeks of treatment with a very-low-calorie diet: (05 Marks)
Before After
85 86
95 90
75 72
110 100
81 75
92 88
83 83
94 93
88 82
105 99
We wish to know if these data provide sufficient evidence to allow us to conclude that the treatment is effective in causing weight reduction in obese women at . Solve using five step critical - value approach.
Click here to see answer by Boreal(15235) |
Question 1196806: Calculate the p-value of the test to determine that there is sufficient evidence to infer each research objective.
Research objective: The population mean is greater than 0.
σ=10, n=100, x-bar= 1.5
Click here to see answer by Theo(13342) |
Question 1197931: Arnold Palmer and Tiger Woods are two of the best golfers to ever play the game. To show how these two golfers would compare if both were playing at the top of their game, the following sample data provide the results of 18-hole scores during a PGA tournament competition. Palmer’s scores are from his 1960 season, while Woods’s scores are from his 1999 season (Golf Magazine, February 2000).
Palmer, 1960 Woods, 1999
n_1 = 112 n_2 = 84
x_1 = 69.95 x_2 = 69.56
Use the sample results to test the hypothesis of no difference between the population mean 18-hole scores for the two golfers.
a. Assume a population standard deviation of 2.5 for both golfers. What is the value of the test statistic?
b. What is the p-value?
c. At α = .01, what is your conclusion?
Click here to see answer by ewatrrr(24785)  |
Question 1198408: A researcher claims that 15% of children suffer from chronic ear infections. A sample of 100 children was taken and it was found that 18% of the children suffer from chronic ear infections. Test at the 1% significance level if the proportion of children that suffer from a chronic ear infection is different from 15%. What is the p-value?
Click here to see answer by ewatrrr(24785) |
Question 1200159: A school conducts an interest aptitude test for final students, the teacher think that the average student aptitude test score is 470 with a known standard deviation value of 10, so the school takes a sample of the talent interest test value data from 9 students as follows:
Student A Score 440
Student B Score 460
Student C Score 485
Student D Score 509
Student E Score 496
Student F Score 477
Student G Score 457
Student H Score 465
Student I Score 495
The hypothesis put forward by the management department is as follows:
H0: μ = 470
H1: μ # 470
By using a = 5%, testing in two tails (two sides), test this opinion and give the conclusion!
Note:
Do this problem using manual formulas in a structured manner, don't use calculations from excel, minitab, or the like.
Click here to see answer by math_tutor2020(3817) |
Question 1200133: -An experimental surgical procedure is being studied as an alternative to the existing method. Twelve surgeons perform the operation on two different patients matched by sex, age and other relevant factors.
Problem Definition: Determine whether or not the new procedure is faster than the existing procedure. Alpha is .05
I NEED HELP WITH THIS STEP BELOW. I PUT ALL THE DATA INTO MINITAB AND CAME UP WITH THE STATISTICS BUT I'M NOT SURE BOW TO INTERPRET IT:
-Utilize all three techniches: critical value technique, the confidence interval and the p-value in your conclusion. Include the numbers from the minitab output for each technique. Write your assumption and discuss how you know whether or not normality may be assumed.
Descriptive Statistics:
Sample N Mean StDev SE Mean
New Procedure 12 13.25 2.38 0.69
Old Procedure 12 23.42 3.58 1.03
Estimation for Paired Difference:
Mean StDev SE Mean 95% Upper Bound
for μ_difference
-10.17 3.83 1.11 -8.18
µ_difference: population mean of (New Procedure - Old Procedure)
Test:
Null hypothesis H₀: μ_difference = 0
Alternative hypothesis H₁: μ_difference < 0
T-Value P-Value
-9.19 0.000
Click here to see answer by math_tutor2020(3817) |
Question 1201813: In a study of 420,113 cell phone users, 136 subjects developed cancer of the brain or nervous system. Test the claim of a somewhat common belief that such cancers are affected by cell phone use. That is, test the claim that cell phone users develop cancer of the brain or nervous system 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 0.001significance level. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method and the normal distribution as an approximation to the binomial distribution.
Click here to see answer by Theo(13342) |
Question 1205109: Ms. Lisa Monnin is the budget director for the New Process Company. She would like to compare the daily travel expenses for the sales staff and the audit staff. She collected the following sample information:
Sales ($)
131
135
146
165
136
142
Audit ($)
130
102
129
143
149
120
139
At the 0.10 significance level, can she conclude that the mean daily expenses are greater for the sales staff than the audit staff?
a. State the decision rule. (Round the final answer to 3 decimal places.)
Reject H0 if t >
1.363
.
b. Compute the pooled estimate of the population variance. (Round the final answer to 2 decimal places.)
Pooled estimate of the population variance
c. Compute the value of the test statistic. (Round the final answer to 3 decimal places.)
Value of the test statistic
d. State your decision about the null hypothesis.
Reject
H0 : μs ≤ μa.
e. Estimate the p-value. (Round the final answer to 4 decimal places.)
The p-value is
.
Click here to see answer by Theo(13342) |
Question 1205253: A research firm conducted a survey to determine the mean amount smokers spend on cigarette during a week. A sample of 49 smokers revealed that the sample mean is Br. 20 with standard deviation of Br. 5. Construct 95% confidence interval for the mean amount spent.
Click here to see answer by math_tutor2020(3817) |
Question 1205251: A quality controller of a company plans to inspect the average diameter of small bolts made. A random sample of 6 bolts was selected. The sample is computed to be 2.0016mm and the sample standard deviation 0.0012mm. Construct the 99% confidence interval for all bolts made.
Click here to see answer by Theo(13342) |
Question 1205357: A survey of 300 students at York University revealed that 112 favour an NDP candidate for M.P. Construct a 98% confidence interval for the percentage of all York University students who favour an NDP candidate.
Click here to see answer by Theo(13342) |
Question 1198711: Please help me solve this:
Test the indicated claim about the means of two populations. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal.
Women: xbar1=12.5hrs, s1=3.9hrs, n1=14
Men: xbar2=13.8hrs, s2=5.2hrs, n2=17
Use a 0.05 significance level to test the claim that the mean amount of time spent watching television by women is smaller than the mean amount of time spent watching television by men.
Include(a) the null and alternative hypotheses, (b) test statistic, (c) critical value(s) or P-value (or range of P-value) as appropriate, (d) conclusion about the null hypotheses, and (e) conclusion about the claim in your answer.
Note: Value of Test statistic t≈-0.795
Click here to see answer by textot(100) |
Question 1198706: Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Use either the tradition method or P-value method. Identify (a) the null and alternative hypotheses, (b) test statistic, (c) critical value(s) or P-value (or range of P-value) as appropriate, and (d) state the final conclusion that addresses the final claim.
In tests of a computer component, it is found that the mean time between failures is 520 hours. A modification is made which is supposed to increase the time between failures. Tests on a random sample of 10 modified components resulted in the following times (in hours) between failures.
518 548 561 523 536
499 538 557 528 563
At the 0.05 significance level, test the claim that for the modified components, the mean time between failures is greater than 520 hours? You may assume the times between failures are normally distributed. Use the P-value method of testing hypotheses.
Note: The value of sample mean (x̄≈537.1hrs) and sample standard deviation (s≈20.701hrs) are calculated.
Click here to see answer by textot(100) |
Question 1193449: A paired difference experiment produced the following results:
nD=49, x1= 176, x2= 183, xD=−7, sD=58,
(a) Determine the rejection region for the hypothesis H0:μD=0 if Ha:μD>0. Use α=0.03
- t> ____. (5 sig. figs)
(b) Conduct a paired difference test described above.
- The test statistic is ______. (5 sig. figs)
Click here to see answer by yurtman(42) |
Question 1185744: A researcher is interested in the conditions that affect the number of dreams per month that people report in which they are alone. We will assume that based on extensive previous research, it is known that in the general population the number of such dreams per month follows a normal curve, with Population M = 5 and Population SD = 4. The researcher wants to test the prediction that the number of such dreams will be greater among people who have recently experienced a traumatic event. Thus, the researcher studies 36 individuals who have recently experienced a traumatic event, having them keep a record of their dreams for a month. Their mean number of alone dreams is 8. SHould you conclude that people who have recently had a traumatic experience have a significantly different number of dreams in which they are alone? (A) Carry out a Z test using the five steps of hypothesis testing (use the .05 significance level) Please include the Population M, Population SD, N, M, one tail or two tail, Population Mm, Population SD2m, Standard Deviation population SDm, Null Hypothesis
Click here to see answer by CPhill(1959)  |
Question 1185740: a government sponsored telephone counseling service for adolescents tested whether the length of calls would be affected by a special telephone system that had a better sound quality.Over the past several years, the legths of telephone calls (in minutes) were normally distributed with Population M = 18, and Population SD = 8. They arranged to have the special phone system loaned to them for one day. On that day, the mean length of the 46 calls they received was 21 minutes. Test whether the length of calls has changed using the .05 signifance level (A) Carry out the Z-test using the 5 steps of hypothesis testing. Please include the Pop M, Pop SD, N, M, one tail or two tail, Mean Population Mm, Population SD2m, Population SDm
Click here to see answer by CPhill(1959) |
Question 1181434: A random sample of 250 bottles of juice drink were taken and was
found to have an average content that is fess than the company’s claim
that each bottle contains 500 mL of juice drink. Suppose that an
appropriate test statistic revealed a value of -1.75, is there enough
evidence to support their claim at 95% confidence? Sketch the rejection
region and locate test statistic value.
Click here to see answer by CPhill(1959) |
Question 1181285: A recent gasoline survey said that the national average price of gasoline was $1.498 a gallon It was felt that gasoline in Texas was significantly lower than the national averageA survey of 10 different suburbs in Dallas , Texas found the average price of gasoline to be a gallon with a standard deviation of $0.326. Find the p-value for this hypothesis test.
Click here to see answer by CPhill(1959) |
Question 1179924: A car salesman claims that the variance of prices on convertibles is higher than the variance of prices on station wagons. The standard deviation of the list price on 16 convertibles is $6800 and the standard deviation on 24 station wagons is $3900. What should the test value be?
Click here to see answer by CPhill(1959) |
Question 1177872: A poll was taken of 100 college freshmen to determine whether chocolate and vanilla was the preferred ice cream flavor. It was expected that 25% would prefer chocolate and 75% would prefer vanilla. Lets look at the results of the survey. Instead of 25% of the students polled preferring chocolate, it turns out 39% preferred it,and instead 75% of the students polled preferring vanilla,it turned out that 61% preferred it.
a)Compute x2 (chi-square)
b) state and explain the conditions necessary for the application of X2 (chi-square)
Click here to see answer by CPhill(1959) |
Question 1177326: A well-known consulting firm wants to test how it can influence the proportion of questionnaires returns for its surveys. In the belief that the inclusion of an inducement to respond may be influential, the firm sends out 1000 questioners: 200 promise to send respondents a summary of the survey results; 300 indicate that 20 respondents (selected by a lottery) will be awarded gifts; and 500 are accompanied by no inducements. Of these, 80 questionnaires promising a summary, 100 questionnaires offering gifts, and 120 questionnaires offering no inducements are returned. What can you conclude from these results?
Click here to see answer by CPhill(1959) |
Question 1177144: The policy of Banko Metro is that its ATM’s must be stocked with enough cash to satisfy customers
making withdrawals over an entire weekend. An analysis of all withdrawals from two branches is shown in
the table below. Test at α = 0. 01 significance
Branch Mean μ Standard Deviation θ N
A Php6,800.00 Php1,200.00 2500
B Php6,790.00 Php1,400.00 2000
Click here to see answer by CPhill(1959) |
Question 1171313: Inference (Two Populations), Chi-Squared Tests
1. A study comparing children’s reading age (in months) was developed using identical twin
toddlers. One set of 6 twins played for 2 hours each day with educational toys
(experimental group), the corresponding set of 6 twins played for 2 hours each day with
non-educational toys (control group). The mean difference in reading age between the
experimental group and control group of 6 sets of twins was -2.33 months and the
standard deviation of the sample difference was 2.16 months. Set up a hypothesis test to
determine if there is a difference between the 2 groups and use the appropriate sample
test statistic to determine if the difference in reading age is significant at the 5% level
Click here to see answer by CPhill(1959) |
Question 1170647: Dr. James has initiated work on cancer systems biology. He has obtained 3 different cell lines for the expression (in nM) of five proteins. Here is the data that he got.
Proteins
Cell lines P53 Akt Cyclin D mTor GLUT
Colon 130 121 160 131 171
Pancreatic 123 113 158 106 165
Lung 121 112 164 102 149
At the 0.01 and 0.05 level of significance, is there a significant difference protein expression among?
(a)Proteins? (b) Cell lines?
Click here to see answer by CPhill(1959) |
Question 1170626: - Time Series and Non-Parametric Tests
1. The number of fishing rods selling each day is given below. Perform analyses of the time series to determine which model should be used for forecasting. (10 points)
a. 3 day moving average analysis
b. 4 day moving average analysis
c. 3 day weighted moving average analysis with weights w1=0.2, w2=0.3 and w3=0.5 with w1 on the oldest data
d. exponential smoothing analysis with a = 0.3.
e. Which model provides a better fit of the data?
f. Forecast day 13 sales of fishing rods using the model chosen in part (e).
day number of rods sold
1 60
2 70
3 110
4 80
5 70
6 85
7 115
8 105
9 65
10 75
11 95
12 85
(time series and non parametric test)
Click here to see answer by CPhill(1959) |
Question 1170624: A manager at the head office of a company is considering 3 branch office managers for promotion. Branch reports include records of sales volume per agent for each branch. A random sample of records was selected for agents at each branch. All branches are located in cities with similar demographics (per capita income, population etc.) Using the samples, the manager wants to see if there is a significant difference in performance of agents at the three branches. If there is a difference, the information will be used to help determine which branch manager to promote; otherwise it will not be included in the decision. (All values are in hundreds of thousands of dollars.)
branch managed by Harrison: 7.2 ,6.4, 10.1, 11, 9.9, 10.6
branch managed by Dale: 8.8, 10.7, 11.1, 9.8
branch managed by Stevenson: 6.9, 8.7, 10.5, 11.4
For the single measurement problem, use an a = 0.01 level of significance. Conduct an appropriate hypothesis test and conclude whether to reject or not reject the claim that there is no difference among the agents at the different branches.
Click here to see answer by CPhill(1959) |
Question 1170375: 2) The paired data below consist of the test scores of 6 randomly selected students and the number of hours they studied for the test. Use a 0.05 significance level to test the claim that there is a linear correlation between hours studied and test score.
Hours 5 10 4 6 10 9
Scores 64 86 69 86 59 87
1) Null and Alternative Hypothesis
2) Calculator Work
3) Test Statistic, P-Value and Correlation Coefficient r, r=0.2242
4) Conclusion about the null hypothesis
Method 1:
Method 2:
5) Final conclusion that addresses the original claim
Click here to see answer by CPhill(1959) |
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