Construct a 98% confidence interval on the population standard deviation speedup of CNNs.
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An engineer considered the speedup of cellular neural networks (CNNs) for a parallel general purpose computing architecture based on seven transputers in different areas. The speedup data of CNNs for seven areas as in Table 3 is collected to investigate the effectiveness of those transputers. Table 3
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3.7753 |
3.3507 |
4.2180 |
4.0303 |
4.6400 |
4.2681 |
4.8243 |
Construct a 98% confidence interval on the population standard deviation speedup of CNNs.
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- From the Behavioral Risk Factor Surveillance System in 2019, the prevalence of self- reported obesity among Michigan adult residents was 35.0%. Moreover, national reports from American Heart Association and Mayo Clinic state that people are more inactive than they used to because of the COVID-19 preventive measures. With this, you are hypothesizing that this year, the proportion of people categorized as obese is higher than before. To verify this, a random sample of 245 individuals were randomly selected from the different counties. Out of the sample, 98 of them were classified as obese. Using 10% level of significance, perform a test of significance.A research team is interested in investigating the effectiveness of a new medication intended to reduce resting heart rate (beats per minute, BPM) in overweight patients. In order to test the medication, the researcher organizes a clinical trial with participants randomly assigned to receive either the new medication or a placebo. Suppose the following table represents the heart rates for a SRS from both groups. Data for the placebo group are {71, 71, 68, 67, 68, 69, 70, 70, 69, 72, 73} Data for the treatment group are {67, 66, 64, 64, 65, 64, 59, 63, 71, 64} Sample standard deviation for the placebo group is: 1.834 Sample standard deviation for the treatment group is: 3.057 What is the confidence interval? (alpha=0.05) Group of answer choices (0.969492 to 9.266508) (2.31312 to 7.92288) (2.598132 to 7.637868) (-1.796934 to 12.032934)The average height of all male students in ENG2015 in section A is 5 feet and 9 inches (or 175.26 centimeters). A sample of 18 male is selected, and their heights (in centimeters) are in the table below. Develop a 95% confidence interval to estimate the population mean.
- A research team is interested in investigating the effectiveness of a new medication intended to reduce resting heart rate (beats per minute, BPM) in overweight patients. In order to test the medication, the researcher organizes a clinical trial with participants randomly assigned to receive either the new medication or a placebo. Suppose the following table represents the heart rates for a SRS from both groups. Data for the placebo group are {71, 71, 68, 67, 68, 69, 70, 70, 69, 72, 73} Data for the treatment group are {67, 66, 64, 64, 65, 64, 59, 63, 71, 64} Sample standard deviation for the placebo group is: 1.834 Sample standard deviation for the treatment group is: 3.057 What is the Standard Error of the Mean Difference? (alpha=0.05) Group of answer choices 1.834 1.240 1.114 3.057A research team is interested in investigating the effectiveness of a new medication intended to reduce resting heart rate (beats per minute, BPM) in overweight patients. In order to test the medication, the researcher organizes a clinical trial with participants randomly assigned to receive either the new medication or a placebo. Suppose the following table represents the heart rates for a SRS from both groups. Data for the placebo group are {71, 71, 68, 67, 68, 69, 70, 70, 69, 72, 73} Data for the treatment group are {67, 66, 64, 64, 65, 64, 59, 63, 71, 64} Sample standard deviation for the placebo group is: 1.834 Sample standard deviation for the treatment group is: 3.057 What is the test statistics (tstat)? (alpha=0.05) Group of answer choices 4.127 2.791 1.674 4.594A 2012 survey of 417 American adults indicates that 15% of cell phone owners do their browsing on their phone rather than a computer or other device. According to an online article, a report from a mobile research company indicates that 38% of Chinese mobile web users only access the internet through their cell phones. We wish to conduct a hypothesis test to determine if these data provide strong evidence that the proportion of Americans who only use their cell phones to access the internet is less than the Chinese proportion of 38%. (a) The Null Hypothesis, in mathematical symbols, is: H0: p = ______ (b) The Alternative Hypothesis, in mathematical symbols, is: HA : p = _______ In both the null and alternative hypotheses, the symbol ( p ) is the proportion of American adult cell phone owners who do their browsing on their phone. (c) The standard error based on the null hypothesis is... :________ (d) Our data indicates a proportion of Americans who use their phones to browse is…
- Researchers want to test a new hand exercise plan for arthritis patients to see if they have improved flexibility. They recruited 110 patients with arthritis in their hands and randomly assigned them to one of two treatment groups: group i took their regular medicine and did the new hand exercises; and group 2 only took medicine. After one month, 42 of the 55 in group 1 improved in their flexibility, and 37 of the 55 in group 2 improved their flexibility. Based on the 95% confidence interval (-0.076, 0.258), is there convincing evidence of a difference in the proportion of patients like these who would show improvement in flexibility if they did the new hand exercises along with taking the medicine versus just taking the medicine? v to believe there va difference in the proportion of patients like these who would show improvement in flexibility if they did the new hand exercises along with taking the medicine versus just taking the medicine. Because v contained in the interval, there…A job candidate at a large job fair can be classified as unacceptable, provisional, or acceptable. Based on past experience, a high-quality candidate is expected to get 80 percent acceptable ratings, 15 percent provisional ratings, and 5 percent unacceptable ratings. A high-quality candidate was evaluated by 100 companies and received 60 acceptable, 25 provisional, and 15 unacceptable ratings. A chi-square goodness-of-fit-test was conducted to investigate whether the evaluation of the candidate is consistent with past experience. What is the value of the chi-square test statistic and number of degrees of freedom for the test? x² (15–5)° (25–15) (60–80)² with 2df || 15 80 (15–5) (25–15) (60–80)? B with 3df 5 15 80 (15–5)² (25–15) (60–80)² with 99df 15 80 x² (5–15) (15–25) (80–60)² D with 2df 15 25 60 (5–15) (15–25)² (80–60)? E x² with 3df 15 25 60 ||A pharmaceutical company has developed a drug that is expected to reduce hunger. To test the drug, three samples of rats are selected with n = 11 in each sample. The first sample receives the drug every day. The second sample is given the drug ance a week, and the third sample receives no drug at all (the control group). The dependent variables is the amount of food eaten by each rat aver a 1-month period. These data are analyzed by an ANOVA, and the results are reported in the following summary Lable. Fill in all missing values in the Lable. (Hint: Start with the df column.) s.S. d.f. M.S. IF Between 6.57 Within 4.35 TOTAL Use the =FDIST () function in Excel to locate the p-value for this ANOVA: pvalue- Report answer accurate to at least 4 decimal places. If you use a signilficance level of a = 05, what would you condlude about these treatments? OThese data do not pravide evidence of a difference between Lhe treatments OThere is a significant difference between treatments
- Do less agreeable Facebook users tend to spend more time on Facebook? A study (Ivcevic and Ambady, 2013) aimed to investigate various facets of use of the social networking website Facebook. Previous research suggested that individuals with poorer face-to-face social skills may find internet-based social platforms less daunting, so as part of the study Ivcevic and Ambady investigated whether an individual's volume of Facebook use was associated with levels of "agreeableness". The subjects were 99 university students who all had a Facebook page. The subjects agreed to allowing the content of their Facebook pages to be downloaded on six separate occasions over a three-week period. As part of the data gathered on each subject, each of the 99 were individually rated by a selected friend on an "agreeableness" scale (based on average ratings on a questionnaire where all responses were on a Likert scale), with 1 being "low" and 5 "high". We consider here the relationship between "friend…Consider the least-squares regression line Cognition = 85.3 – 2.54 Years %3D that predicts the score on a cognition test given the number of years playing hockey. Find the predicted cognition score for the given case. Two intervals are shown below: one is a 95% confidence interval for the mean response and the other is a 95% prediction interval for the response. Which is the 95% confidence interval for the mean response? Given a person who has played hockey for 5 years: Interval I: (28.2, 117.0) and Interval lI: (62.2, 83.0) O The predicted cognitive score is 72.60 for a person who has played hockey for 5 years. The 95% confidence interval for the mean response is ( 28.2, 117.0). O The predicted cognitive score is 72.60 for a person who has played hockey for 5 years. The 95% confidence interval for the mean response is (62.2, 83.0). O The predicted cognitive score is 85.3 for a person who has played hockey for 5 years. The 95% confidence interval for the mean response is ( 28.2,…The Wall Street Journal asked Concur Technologies, Inc., an expense management company, to examine data from 8.3 million expense reports to provide insights regarding business travel expenses. Their analysis of the data showed that New York was the most expensive city. The following table shows the average daily hotel room rate (X) and the average amount spent on entertainment (Y) for a random sample of 9 of the 25 most-visited U.S. cities. These data lead to the estimated regression equation y = 17.49 + 1.0334x. For these data SSE = 1541.4. Use Table 1 of Appendix B. a. Predict the amount spent on entertainment for a particular city that has a daily room rate of $89 (to 2 decimals). b. Develop a 95% confidence interval for the mean amount spent on entertainment for all cities that have a daily room rate of $89 (to 2 decimals). c. The average room rate in Chicago is $128. Develop a 95% prediction interval for the amount spent on entertainment in Chicago (to 2 decimals).
