High Cholesterol: A group of eight individuals with high cholesterol levels were given a new drug that was designed to lowercholesterol levels. Cholesterol levels, in milligrams per deciliter, were measured before and after treatment for each individual,with the following results:Individual Before After124721122812173.26019042762145.24015662802142931958274181Send data to ExcelPart: 0/2Part 1 of 2(a) Construct an 80% confidence interval for the mean reduction in cholesterol level. Let d represent the cholesterol levelbefore treatment minus the cholesterol level after. Use tables to find the critical value and round the answers to at least onedecimal place.An 80% confidence interval for the mean reduction in cholesterol level is

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Mathematics For Machine Technology

8th Edition

ISBN:9781337798310

Author:Peterson, John.

Publisher:Peterson, John.

Chapter29: Tolerance, Clearance, And Interference

Section: Chapter Questions

Problem 16A: Spacers are manufactured to the mean dimension and tolerance shown in Figure 29-12. An inspector...

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Spacers are manufactured to the mean dimension and tolerance shown in Figure 29-12. An inspector measures 10 spacers and records the following thicknesses: 0.372" 0.376" 0.379" 0.375" 0.370" 0.373" 0.377" 0.378" 0.371" 0.380" Which spacers are defective (above the maximum limit or below the minimum limit)? All dimensions are in inches.
Find the mean hourly cost when the cell phone described above is used for 240 minutes.
What are the missing values for Between and Within treatment, in terms of DF, MS and F?
Glaucoma is a leading cause of blindness in the United States, N. Ehlers measured the difference in corneal thickness (in microns) between the two eyes of eight patients. Each patient had one eye that had glaucoma and one eye that was normal. The difference was measured as the corneal thickness of normal eye – corneal thickness of eye with Glaucoma. Corneal thickness is important because it can mask an accurate reading of eye pressure. Use ? = .05 H0: μd=0 H1: μd≠0 T statistic: t = 1.053, P value = 0.327 Degrees of Freedom n-1 = 8-1= 7 critical value 2.3646 If t is less than 2.3646, or greater than 2.3646, reject the null hypothesis Level of significance: α=0.05 question: what a type 1 error and type 2 error would mean. Is it possible that we could have committed a type 2 error in conducting the test
Glaucoma is a leading cause of blindness in the United States, N. Ehlers measured the difference in corneal thickness (in microns) between the two eyes of eight patients. Each patient had one eye that had glaucoma and one eye that was normal. The difference was measured as the corneal thickness of normal eye – corneal thickness of eye with Glaucoma. Corneal thickness is important because it can mask an accurate reading of eye pressure. Use ? = .05. Q) If a participant has the same corneal thickness in their normal eye as the eye with Glaucoma, what would be the value for difference: measured as the corneal thickness of normal eye – corneal thickness of eye with Glaucoma.
Glaucoma is a leading cause of blindness in the United States, N. Ehlers measured the difference in corneal thickness (in microns) between the two eyes of eight patients. Each patient had one eye that had glaucoma and one eye that was normal. The difference was measured as the corneal thickness of normal eye – corneal thickness of eye with Glaucoma. Corneal thickness is important because it can mask an accurate reading of eye pressure. Use ? = .05. Q)Conduct a hypothesis test to determine if there is sufficient evidence to conclude that corneal thickness is different in normal eyes compared to eyes with glaucoma? Write up your results using the 8 steps.
Glaucoma is a leading cause of blindness in the United States, N. Ehlers measured the difference in corneal thickness (in microns) between the two eyes of eight patients. Each patient had one eye that had glaucoma and one eye that was normal. The difference was measured as the corneal thickness of normal eye – corneal thickness of eye with Glaucoma. Corneal thickness is important because it can mask an accurate reading of eye pressure. Use ? = .05. Hypothesis: H0: μd=0Ha: μd≠0 Using output Test statistics : t=0.134P value=0.897 Degrees of freedom (df): df=7 Level of significance: α=0.05 Decision: P value > 0.05 thus we fails to reject null hypothesis. Question a)Write a report summarizing your findings. When writing the report consider that medical staff estimate that a difference of 4.5 microns or more could impact on their ability to interpret eye pressure correctly. b) Define for the hypothesis stated in part b) what a type 1 error and type 2 error would mean. Is it possible…
Refer to the table to the right summarizing measured amounts of serum cotinine (ng/mL) from a sample of smokers. When nicotine is absorbed by the body, cotinine is produced. How many subjects are included in the summary? Is it possible to identify the exact values of all of the original cotinine measurements? Cotinine (ng/mL) 0-99 100-199 200-299 300-399 400-499 How many subjects are included in the summary? (Type a whole number.) Is it possible to identify the exact values of all of the original cotinine measurements? Frequency 10 14 15 3 2 OA. Yes. The exact data values are equally spaced numbers between the limits of each class. The number of equally spaced numbers is determined by the frequency in each class. OB. No. A frequency table only shows percentages of the sample in each category. OC. No. A frequency table only shows frequencies of data values in each category. OD. Yes. The data values are the frequencies shown in the table.
Periodically, the county Water Department tests the drinking water of homeowners for contaminants such as lead and copper. The lead and copper levels in water specimens collected in 1998 for a sample of 10 residents of a subdevelopement of the county are shown below. lead (g/L) 2.9 0.2 5.1 4.2 5.5 1.2 0.133 0.774 0.214 0.671 0.444 0.234 0.357 0.761 0.176 0.888 (a) Construct a 99% confidence interval for the mean lead level in water specimans of the subdevelopment. OSMOSO copper (mg/L) 0.3 1.3 4.9 1.7 (b) Construct a 99% confidence interval for the mean copper level in water specimans of the subdevelopment. OSHSO
An abrasion test was performed using n = 14 moulded ceramic pieces. The following are the test results, where for the i-th piece, the result of the test was recorded (one of Survived, Broken). The data also gives Xi, the porosity index of the piece, which is a numerical quantity: (Survived,8), (Broken,16), (Broken,19), (Survived,26), (Survived,49), (Survived,54), (Survived,66), (Survived,79), (Survived,102), (Broken,111), (Survived,114), (Broken,151), (Survived,163), (Broken,199) We want to study the resistance of pieces to abrasion as function of porosity. To this end, consider the following logistic regression model, log πi 1/727 ) = B₁ + B₁x₁₂ πi where the observations follow the distribution Y; Bernoulli(;). A data value y; = 1 indicates that the piece survived the test and yi 0 indicates that the piece broke. Fit this model in R using the function glm (). Examine your output and then answer the following questions. = A) What is the value of Bo? B) If the porosity of a piece was…
How sensitive to changes in water temperature are coral reefs? To find out, scientists examined data on sea surface temperatures and coral growth per year at locations in the Gulf of Mexico and the Caribbean Sea. Scientists examined data on mean sea surface temperatures (in degrees Celsius) and mean coral growth (in centimeters per year) over a several‑year period at locations in the Gulf of Mexico and the Caribbean. The table shows the data for the Gulf of Mexico. Sea surface temperature 26.726.7 26.626.6 26.6 26.526.5 26.326.3 26.126.1 Growth 0.850.85 0.850.85 0.790.79 0.860.86 0.890.89 0.920.92 (a) Use software to find the mean and standard deviation of both sea surface temperature ?x and growth ?y and then find the correlation ?r between ?x and ?.y. Enter your answers in the answer fields using the specific decimal place precision noted. ?¯ (rounded to four decimal places)=x¯ (rounded to four decimal places)= ?? (rounded to four decimal places)=sx…
A large automobile insurance company selected samples of single and married male policyholders and recorded the number who made an insurance claim over the preceding three-year period. Single Policyholders Married Policyholders 350 1,000 Number making claims = 70 Number making claims = 150 a. Use l= 0.05 Test to determine whether the claim rates differ between single and married male policyholders. z-value (to 2 decimals) p-value (to 4 decimals) b. Provide a 95% confidence interval (to 4 decimals) for the difference between the proportions for the two populations. Enter negative answer as negative number.

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