Remember that in Pisum sativum, yellow seed color is dominant over green, and smooth seed is dominant over constricted. In a dihybrid cross experiment, the following data was gathered: 189 green, smooth 178 yellow, constricted 564 yellow, smooth 69 green, constricted Are these genes assort independently based on the observed data? Support your answer using Chi-square test. Ans.
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- Suppose a researcher conducted a study examining the effectiveness of font type (OpenDyslexic vs. Times New Roman) on reading comprehension for 20 students with dyslexia. After reading a passage in one of the two fonts, 10 students in each of the two font conditions completed a brief reading comprehension test of 15 questions. The number of correctly answered questions for each student is given in the table below: Complete the following in the context of conducting data analysis/ a. Calculate the average number of correct test questions for each group b. Calculate the standard deviation for each group c. Calculate the median and mode for both groups d. Calculate the interquartile range for each group OpenDyslexic 10 9 11 8 24867 12 14 Times New Roman 6675 & Na 43 4 2 8 9The counts of the F1offspring are listed in Table 1. Two randomly selected individuals are selected and mated to produce a set of F2individuals. If you test the F2counts to determine whether they are consistent with an autosomal dominant mode of inheritance, what is your decision? Use a 0.05 significance level. PARENTAL CROSS Parental cross: Mother with disease phenotype, Father with wild-type phenotype. Table 1. F1 DATA Gender Phenotype Disease Wild-type Male 0 23 Female 0 34 Table 2. F2 DATA Gender Phenotype Disease Wild-type Male 7 25 Female 0 25 A. Do not reject the null hypothesis that the F2 data are consistent with an autosomal dominant mode of inheritance; chi-square goodness of fit test p-value is greater than 0.05. B. Reject the null hypothesis that the F2 data are consistent with an autosomal dominant mode of inheritance; chi-square goodness of fit test p-value is 1e-06. C. Reject the null hypothesis that the F2 data are consistent…An article included data from a survey of 2500 hiring managers and human resource professionals. The article noted that many employers are using social networks to screen job applicants and that this practice is becoming more common. Of the 2500 people who participated in the survey, 1250 indicated that they use social networking sites (such as Facebook, MySpace, and LinkedIn) to research job applicants. For the purposes of this exercise, assume that the sample is representative of hiring managers and human resource professionals -Why can this sampling distribution be considered normal? Explain and show why. -Construct a 95% confidence interval for the true population proportion of employers who use social networking sites to screen job applicants. Be sure to write a conclusion sentence that interprets your interval. (You may use Excel for this -Construct a 99% confidence interval for the true population proportion of employers who use social networking sites to screen job applicants.…
- (Non Parametric Test) Following are the weight gains in pounds of 2 randomly samples of young turkeys fed 2 different diets: Diet 1: 16.3, 10.1, 10.7, 13.5, 14.9, 11.8, 14.3, 10.2, 12.0, 14.7, 23.6, 15.1, 14.5, 18.4, 13.2 & 14.0 Diet 2: 21.3, 23.8, 15.4, 19.6, 12.0, 13.9, 18.8, 19.2, 15.3, 20.1, 14.8, 18.9, 20.7, 21.1, 15.8, & 16.2 Test @ 0.01 level of significance to test Ho that the 2 population sampled are identical against HA that on the average the second diet produces a greater gain in weight.Two autosomal loci control horn color in unicorns. Pure-breeding gold-horned unicorns were mated to pure-breeding bronze unicorns. The F1 were all silver. The F2 generation consisted of 9 gold, 7 bronze, and 22 silver unicorns. Use these data to hypothesize the type of epistasis that is operating here. Ho: The mode of inheritance for horn color in unicorns is A larger experiment was conducted to test the null hypothesis. The F2 generation in the larger data set consisted of 126 bronze-horned unicorns, 166 gold horned unicorns and 392 silver horned unicorns. Conduct the Chi-square test using the larger data set. Round all expected numbers correctly to three decimal digits. The expected number of Gold Unicorns is The expected number of Silver Unicorns is The expected number of Bronze Unicorns is Rounded to three decimal digits, the calculated chi-square value is The number of degrees of freedom for this test is Rounded to two decimal digits, the critical value is The statistical…A study indicates that CB radio use speeds up activity in the brain. As part of a randomized crossover study, 36 participants were fitted with a CB radio device on each ear and then underwent two positron emission topography, or PET, scans to measure brain glucose metabolism. During one scan, the CB radios were both turned off. During the other scan, an automated 40- minute muted signal was sent to the radio on the right ear. The order of when the signal was received (for the first or second scan) was randomized. Comparison of the PET scans showed a minor increase in activity in the part of the brain closest to the antenna during the transmission of the automated signal. Was this an observational study or an experiment? Explain why. O The study is an observational study because the reseashers cannot control the response variable during the signal. O The study is an observational study because the researchers only observe values of the response and explanatory variables. O The study is…
- Hughes randomly samples 100 people and shows both menus to each person, asking them to rate each menu from 0 (very poor) to 20 (excellent) and then wonders whether the difference is statistically significant. [ Select ]The dry shear strength of birch plywood bonded with different resin glues was studied with a completely randomized designed experiment. Here are the data: Glue A (102; 58; 45; 79; 68; 63; 117) Glue C (100; 102; 80; 119) Glue F (220; 243; 189; 176; 176). SSE = 8167.55; MSE = 628.27; and F stat = 37.99. What is the value for the F test statistics (rounded off 2 decimals)?An experimental study was conducted on the effect of programmed materials in English on the performance of 20 selected college students. Before the program was implemented pretest was administered and after 5 months the same instrument was used to get the post-test result.
- A study indicates that CB radio use speeds up activity in the brain. As part of a randomized crossover study, 36 participants were ntted with a CB radio device on each ear and then underwent two positron emission topography, or PET, scans to measure brain glucose metabolism. During one scan, the CB radios were both turned off. During the other scan an automated 40- minute muted signal was sent to the radio on the right ear. The order of when the signal was received (for the first or second scan) was randomized. Comparison of the PET scans showed a minor increase in activity in the part of the brain closest to the antenna during the transmission of the automated signal. Was this an observational study or an experiment? Explain why. O The study is an observational study because the researchers cannot control the response variable during the signal. O The study is an observational study because the researchers only observe values of the response and explanatory variables. O The study is…The news release referenced in the previous exercise also included data fromindependent samples of teenage drivers and parents of teenage drivers. In response to aquestion asking if they approved of laws banning the use of cell phones and texting whiledriving, 74% of the teens surveyed and 95% of the parents surveyed said they approved.The sample sizes were not given in the news release, but suppose that 600 teens and 400 parents of teens were surveyed and that these samples are representative of the twopopulations. Do the data provide convincing evidence that the proportion of teens whoapprove of banning cell phone and texting while driving is less than the proportion ofparents of teens who approve?You conduct a one-factor ANOVA with 6 groups and 5 subjecis in each group (a balanced design) and obtain F = 2.48. Use Excel to find the requested values (report accurate Lo 4 decimal places). dfiuiwmn = dfanhin = P=
