Answered step by step
Verified Expert Solution
Link Copied!

Question

1 Approved Answer

please solve all these questions:- The following sample of six measurements was randomly selected from a normally distributed population: 1, 3. -1, 5, 1, 2.

please solve all these questions:-

image text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribed
The following sample of six measurements was randomly selected from a normally distributed population: 1, 3. -1, 5, 1, 2. a. Test the null hypothesis that the mean of the population is 3 against the alternative hypothesis J 1,000, based on a sample of 17 observations. The test results are f = 1.89, p- value = 038. a. What assumptions are necessary for the validity of this procedure? b. Interpret the results of the test. c. Suppose the alternative hypothesis had been the twotailed Harp # 1,000. If the f-statistic were unchanged, what would the p -value be for this test? Interpret the p -value for the two-tailed test. If the f-statistic were unchanged, what would the p -value be for this test? Interpret the p - value for the two-tailed test.Minimizing tractor skidding distance. Refer to the Journal of Forest Engineering (July 1999) study of minimizing tractor skidding distances along a new road in a European forest. The skidding distances (in meters) were measured at 20 randomly selected road sites. The data (saved in the SKIDDING file) are repeated in the accompanying table. Recall that a logger working on the road claims that the mean skidding distance is at least 425 meters. Is there sufficient evidence to refute this claim? Use a = .10. 488 350 | 457 109 285 409 435 574 439 546 385 205 184 261 273 400 311 |312 141 |425 Based on Tujek, J. and Pacola, E. "Algorithms for skidding distance modeling on a raster Digital Terrain Model." Joumal of Forest Engineering . Vol 10, No. 1. July 1999. Conclusions and Consequences for a Test of Hypothesis True State of Nature Conclusion HO True Ha True Correct decision Accept HO (Assume HO True) Reject HO Type II error (probability 5) (Assume Ha True) Type I error Correct decision (probability a)Dissolved organic compound in lakes. The level of dissolved oxygen in the surface water of a lake is vital to maintaining the lake's ecosystem. Environmentalists from the University of Wisconsin monitored the dissolved oxygen levels over time for a sample of 25 lakes in the state (Aquatic Biology, May 2010). To ensure a representative sample, the environmentalists focused on several lake characteristics, including dissolved organic compound (DOC). The DOC data (measured in grams per cubic meters) for the 25 lakes are listed in the table on p. 400. The population of Wisconsin lakes has a mean DOC value of 15 grams/m . Use a hypothesis test (at " = .10) to make an inference about whether the sample is representative of all Wisconsin lakes for the characteristic dissolved organic compound Data for Exercise 8.70 Lake DOC Allequash 9.6 Hip Muskellunge 4.5 Hrown 13.2 Crampton 4.1 Cranberry Bog 22.6 Crystal 2.7 FasiLong 14.7 Helmet 3.5 Hiawatha 13.6 Hummingbird 19.8 Kickapoo 14.3 Little Arbor Vitae 56.9 Mary 25.1 Muskellunge Northgate Bog 2.7 Paul 4.2 Peter 30.2 Plum 10.3 Reddington Bog 17.6 Sparkling 24 Tenderfoot 17.3 Trout Bog 38.8 Trout Lake 3.0 Ward 5.8 West Long 7.6 Hasad on Langman, Q C, cial "Control of dissolved anygen in northern temperate lakes over wakes ranging from minutes In days" Aquatic Biology, Vol 9, May 2010 (Table 1).Suppose a random sample of 100 observations from a binomial population gives a value of p =.60 and you wish to test the null hypothesis that the population parameter p is equal to.75 against the alternative hypothesis that p is less than.75. a. Noting that p =.80, what does your intuition tell you? Does the value of p appear to contradict the null hypothesis? b. Use the large-sample z-test to test HO: p =.75 against the alternative hypothesis Ha: p<.75. use d=".05." how do the test results compare with your intuitive decision from part a c. find and interpret observed significance level of you conducted in b.suppose sample exercise has produced we wish to ho: p=".9" against alternative ha: a. calculate value z statistic for this test. b. note that numerator p0=".84" is same as considering why absolute larger than calculated complete using result. d. its value.a random observations selected binomial population unknown probability success p. computed equal to.> 65. Use a =.01. b. Test HD: p =.65 against Ha: p >.65. Use a =.10. c. Test HO: p =.90 against Ha: p #.90. Use a =.05. d. Form a 95% confidence interval for p. e. Form a 09% confidence interval for p.In which 50 consumers taste-tested a new snack food. The data are saved in the SNACK file. 3. Test HO: p =.5 against Ha: p >.5, where p is the proportion of customers who do not like the snack food. Use a =.10. b. Report the observed significance level of your test. A random sample of 50 consumers taste-tested a new snack food. Their responses were coded (0: do not like: 1: like: 2: indifferent). recorded and saved in the SNACK file as follows: 1 12 01 1 0 10 2 0 2 2 0 0 1 1 0 0 0 0 1 0 2 0 0 0 1 0 0 1 0 0 1 0 1 0 2 0 0 1 10 1 a. Use an 80%% confidence interval to estimate the proportion of consumers who like the snack food b. Provide a statistical interpretation for the confidence interval you constructed in part a.Dating and disclosure. Refer to the Journal of Adolescence (April 2010) study of adolescents' disclosure of their dating and romantic relationships. Recall that a sample of 222 high school students was recruited to participate in the study. One of the variables of interest was the level of disclosure to an adolescent's mother (measured on a 5-point scale, where 1 = *never tell," 2 ="rarely tell," 3 = "sometimes tell," 4 ="almost always tell," and 5 = "always tell"). The sampled high school students had a mean disclosure score of 3.20 and a standard deviation of 93. The researchers hypothesize that the true mean disclosure score of all adolescents will exceed 3. Do you believe the researchers? Conduct a formal test of hypothesis using o = .01 Dating and disclosure. As an adolescent, did you voluntarily disclose information about dating and romantic relationships to your parents? This was the research question of interest in the Journal of Adolescence (Apr. 2010). A sample of 222 high school students was recruited to participate in the study. Some of the many variables measured on each student were age (years). gender, dating experience (number of dates), and the extent to which the student was willing to tell his/her parent (without being asked) about a dating issue (e.g., how late the daters stayed out). The responses for the last variable were categorized as "never tell," "rarely tell," "sometimes tell," "almost always tell," and "always tell." a. Identify the data type for each variable. b. The study was unclear on exactly how the sample of students was selected, stating only that "participants were recruited from health or government classes in a primarily European American middle-class school district." Based on this information, what are the potential caveats to using the sample to make inferences on dating and disclosure to parents for all high school students?Birth order and IQ. An international team of economists investigated the possible link between 10 and birth order in CESifo Economic Studies (Vol. 57, 2011). The data source for the research was the Medical Birth Registry of Norway. It is known that the mean IQ (measured in stanines) for all Norway residents is 5.2 points. In the study, a sample of 581 Norway residents who were the 6th-born or later in their families had a mean IQ score of 4.7 points with a standard deviation of 1.8 points. Is this sufficient evidence to conclude that the mean IQ score of all Norway residents who were the 6th-born or later in their families is lower than the country mean of 5.2 points? Use w= 01 as a measure of reliability for your inference.Cooling method for gas turbines. During periods of high demand for electricity-especially in the hot summer months-the power output from a gas turbine engine can drop dramatically. One way to counter this drop in power is by cooling the inlet air to the turbine. An increasingly popular cooling method uses high-pressure inlet fogging. The performance of a sample of 87 gas turbines augmented with high-pressure inlet fogging was investigated in the Joumal of Engineering for Gas Turbines and Power (Jan. 2005). One measure of performance is heat rate (kilojoules per kilowatt per hour). Heat rates for the 67 gas turbines, saved in the GASTURBINE file, are listed in the next table. Suppose that a standard gas turbine has, on average, a heat rate of 10,000 kJ/kWh. a. Conduct a test to determine whether the mean heat rate of gas turbines augmented with high- pressure inlet fogging exceeds 10,000 kJ/kWh. Use a =_05. b. Identify a Type I error for this study. Identify a Type II error. 14822 13198 11948 11289 11964 10526 10387 10592 10480 | 10088 14828 13398 11726 11252 12449 11030 10787 10603 10144 11874 11510 10946 10508 10604 10270 |10529 10380 14796 12913 12270 11842 10656 11380 11136 10814 13523 11289 11183 10951 8722 10481 9812 9569 9643 9115 9115 11588 10888 9738 9295 9421 9105 10233 10186 9018 |9209 9532 9933 9152 9295 16243 14628 12786 8714 9469 11948 12414

Step by Step Solution

There are 3 Steps involved in it

Step: 1

blur-text-image

Get Instant Access to Expert-Tailored Solutions

See step-by-step solutions with expert insights and AI powered tools for academic success

Step: 2

blur-text-image

Step: 3

blur-text-image

Ace Your Homework with AI

Get the answers you need in no time with our AI-driven, step-by-step assistance

Get Started

Recommended Textbook for

College Algebra

Authors: Cynthia Y Young

3rd Edition

1118475690, 9781118475690

More Books

Students also viewed these Mathematics questions

Question

6. How can hidden knowledge guide our actions?

Answered: 1 week ago