4.

Construct the cumulative frequency distribution for the given data. Age (years) of Best Actress when award was won Age (years) of Best Actress when award was Cumulative Frequency Frequency won 20-29 28 Less than 30 30-39 33 40-49 11 Less than 40 50-59 Less than 50 60-69 70-79 Less than 60 80-89 Less than 70 Less than 80 " . . Less than 90Find the (a) mean, (b) median, (c) mode, and (d) midrange for the given sample data. An experiment was conducted to determine whether a deciency of carbon dioxide in the soil affects the phenotype of peas. Listed below are the phenotype codes where 1 =smoothryellow, 2 = smoothgreen, 3 = wrinkledryellow, and 4 =wrinkledgreen. Do the results make sense? 2 2 3 3 3 4 3 1 3 3 1 2 2 1 |E (a)The mean phenotype code is |_|. (Round to the nearest tenth as needed.) (b)The median phenotype code is (Type an integer or a decimal.) (c) Select the correct choice below and ll in any answer boxes within your choice. O A- The mode phenotype code is . (Use a comma to separate answers as needed.) 0 B. There is no mode. (d)The midrange of the phenotype codes is (Type an integer or a decimal.) Do the measures of center make sense? 0 A. Only the mean, median, and midrange make sense since the data is nominal. O B. Only the mean, median, and mode make sense since the data is numerical. O C. All the measures of center make sense since the data is numerical. Determine whether the followmg procedure results in a binomial distribution or a distribution that can be treated as binomial (by applying the 5% guideline for cumbersome calculations}. If it is not binomial and cannot be treated as binomial. identify at least one requirement that is not satised. In a Pew Research Center sunreiir of 50 subjects, the ages of the respondents are recorded. Choose the correct answer below. O A. It is binomial or can be treated as binomial. O B. It is not binomial because the probability of success does not remain the same in all trials. C. It is not binomial because there are more than two possible outcomes and the trials are not independent. O D. It is not binomial because there are more than two possible outcomes. Find the area of the shaded region. The graph to the n'ghtdepicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. 106 132 '09 The area of the shaded region is _ (Round to four decimal places as needed.) Conduct the hypothesis test and provide the test statistic and the critical value, and state the conclusion. A person drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed frequencies for the outcomes of 1, 2, 3, 4, 5, and 6, respectively: 25, 31, 47, 41, 26, 30. Use a 0.01 significance level to test the claim that the outcomes are not equally likely. Does it appear that the loaded die behaves differently than a fair die? Click here to view the chi-square distribution table. The test statistic is (Round to three decimal places as needed.) The critical value is (Round to three decimal places as needed.) State the conclusion. Ho. There sufficient evidence to support the claim that the outcomes are not equally likely. The outcomes to be equally likely, so the loaded die to behave differently from a fair die.A data set includes the counts of chocolate chips from three different types of Chips Ahoy cookies. The accompanying StatCrunch display shows results from analysis of variance used with those three types of cookies. Use a 0.05 signifcance level to test the claim that the three different types of cookies have the same mean number of chocolate chips. AN OVA table Source DF 35 MS F-Stat Columns 2 10655500 532.??500 58.5899 Error F0013333 9.0932900 Total 1?65.?333 Determine the null hypothesis. HO: Y Determine the alternative hypothesis. H1: 1' Determine the test statistic. The test statistic is . (Round to two decimal places as needed.) Determine the Pvalue. The Pvalue is (Round to three decimal places as needed ) Identify the level of measurement of the data, and explain what is wrong with the given calculation. In a set ofdata, movie ratings are represented as 100 for 1 star, 200 for 2 stars, and 300 for 3 stars. The average (mean) of the 799 movie ratings is 231 2. The data are at the Y level of measurement. What is wrong with the given calculation? 0 A. One must use a different method to compute the average (mean) of such data. 0 B. Such data should not be used for calculations such as an average (mean). 0 C. The true average (mean) is 195.4 0 D. There is nothing wrong with the given calculation. Heights (cm) and weights (kg) are measured for 100 randomly:r selected adult males, and range from heights of 136 to 191 cm and weights of 38 to 150 kg. Let the predictor variable x be the rst variable given. The 100 paired measurements vie|d=16779 cm, i: 81 46 kg, r= 0.333, Pvalue = 0.001, and y: 103 +1.09x. Find the best predicted value of y (weight) given an adult male who is 151 cm tall. Use a 0.10 Signicance level. The best predicted value of 9 tor an adult male who Is 151 cm tail is kg. (Round to two decimal places as needed.) Here are 6 celebrities with some of the highest net worths (in millions of dollars} in a recent year: George Lucas (5500), Steven Spielberg (BTDD), Oprah Winfrey (3200), J. K. Rowling (1000), David Copperfield (1000), and Jerry Seinfeld (5350) Q1 . Find the range, variance, and standard dewation for the sample data. What do the results tell us about the population of all celebrities? Based on the nature of the amounts, what can be inferred about their precision? The range is $ million. (Round to the nearest integer as needed.) The variance is million dollars squared. (Round to the nearest integer as needed.) The standard deviation is $ million. (Round to the nearest integer as needed.) What do the results tell us about the population of all celebrities? O A. Because the statistics are calculated from the data, the measures of variation are typical for all celebrities. O B. Because the data are from celebrities With the highest net worths, the measures of variation are typical for all celebrities. O (3. Because the data are from celebrities With the highest net worths, the measures of variation are not at all typical for all celebrities. O D. Because the statistics are calcu ated from the data, the measures of variation cannot tell us about other celebrities. Based on the nature of the amounts, what can be inferred about their precision? it appears that they are rounded to the nearest VI so it would make sense to round the resutts to Because all of the amounts end with the nearest Listed below are the numbers of words spoken in a day by each member of eight different randomly selected couples. Complete parts (a) and (b) below Male 15,167 2?,300 1444 ?620 10,504 15,809 14,130 25,485 El Female 23,802 13,12?r 18,33?r 1?,469 12,704 16,522 16,1?8 18,359 In this example, pd is the mean value of the differences d for the population of all pairs of data, where each individual difference d is dened as the words spoken by the male minus words spoken by the female. What are the null and alternative hypotheses for the hypothesis test? HO: pd V word(s) H1: pd V word(s} (Type integers or decimals. Do not round.) Identify the test statistic. t: (Round to two deCimal places as needed ) Identify the Pvalue. Prvalue = (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? Since the 3value is Y the signicance level, V the null hypothesis. There Y sufcient evidence to supportthe claim that males speak fewer words in a day than females. In. Construct the condence interval that could be used for the hypothesis test described in part (a). What feature of the condence interval leads to the same conclusion reached in part (a)? The confidence interval is word(s) = Hd word(s). (Round to the nearest integer as needed.) What feature of the confidence interval leads to the same conclusion reached in part (a)? Since the confidence interval contains the null hypothesis