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In an experiment, college students were given either four quarters or a $1 bill and they could either keep the money or spend it on gum. The results are summarized in the table Complete parts (althrough (c) below. Purchased Gum Kept the Money D; Students Given Four Quarters 2E 13 Students Given a 31 Bill 16 29 3. Find the probability of randomly selecting a student who spent the money, given that the student was given a $1 bill. The probability is . (Round to three decimal places as needed.) b. Find the probability of randomly selecting a student who kept the money, given that the student was given a $1 bill. The probability is . (Round to three decimal places as needed.) c. What do the preceding results suggest? O A. A student given a $1 bill is more likely to have kept the money than a studentgiven four quarters. O B. A student given a $1 bill is more likely to have kept the money. 0 C. A student given a $1 bill is more likely to have spent the money. 0 D. A student given a $1 bill is more likely to have spent the money than a student given four quarters. Here are summary statistics for randomly selected weights of newborn girls: n = 36, x = 3211.1 g, s =689.4 g. Use a confidence level of 99% to complete parts (a) through (d) below. a. Identify the critical value to /2 used for finding the margin of error. to /2= (Round to two decimal places as needed.) b. Find the margin of error. E = 9 (Round to one decimal place as needed.) c. Find the confidence interval estimate of p. (Round to one decimal place as needed.) d. Write a brief statement that interprets the confidence interval. Choose the correct answer below. O A. Approximately 99% of sample mean weights of newborn girls will fall between the lower bound and the upper bound. O B. There is a 99% chance that the true value of the population mean weight of newborn girls will fall between the lower bound and the upper bound. O C. One has 99% confidence that the sample mean weight of newborn girls is equal to the population mean weight of newborn girls. O D. One has 99% confidence that the interval from the lower bound to the upper bound contains the true value of the population mean weight of newborn girls.Assume that the amounts of weight that male college students gain during their freshman year are normally distributed with a mean of p = 1.1 kg and a standard deviation of o = 4.8 kg. Complete parts (a) through (c) below. a. If 1 male college student is randomly selected, find the probability that he gains between 0 kg and 3 kg during freshman year. The probability is (Round to four decimal places as needed.) b. If 25 male college students are randomly selected, find the probability that their mean weight gain during freshman year is between 0 kg and 3 kg. The probability is (Round to four decimal places as needed.) c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30? O A. Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size. O B. Since the distribution is of sample means, not individuals, the distribution is a normal distribution for any sample size. O C. Since the weight gain exceeds 30, the distribution of sample means is a normal distribution for any sample size. O D. Since the distribution is of individuals, not sample means, the distribution is a normal distribution for any sample size.Determine whether the following 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 survey of 20 subjects, the ages of the respondents are recorded. Choose the correct answer below. 0 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. 0 C. It is not binomial because there are more than two possible outcomes. 0 D. It is not binomial because there are more than two possible outcomes and the trials are not independent. Find the area of the shaded region, The graph to the right depicts IQ scores of adults] and those scores are normatly distributed With a mean of 100 and a standard deviation of 15, 102 132 I399 The area of the shaded region is . (Round to tour decimal places as needed.) Find 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 aects the phenotype of peas. Listed below are the phenotype cooes where 1 : smootheyellow' 2: smoothsgreel 3 : wrinkled-yellow, and 4 : wnnkledgreen. Do the results make sense? 3 4 4 a 2 1 4 3 4 1 1 4 4 1 g (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. All the measures of center make sense since the data Is numerical O B. Only the mode makes sense since the data is nominal. O C. Only the mean, median, and mode make sense since the data is numerical. O D. Only the mean, median, and midrange make sense since the data is nominal. 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 significance level to test the claim that the three different types of cookies have the same mean number of chocolate chips. ANOVA table Source DF SS MS F-Stat P-value Columns 2 1149.5500 574.77500 61.9696 H2 ~ H3 Hy = H2 = H3 Hy = H2 =H3 At least one of the means is different from the others All of the means are differentAll of the means are different At least one of the means is different from the others Hy = 12 = H3 H1 > H2 ~ H3 Hy = H2 = H3Here are 6 celebrities with some ofthe highest net worths (in millions of dollars) in a recent year: Oprah Winfrey (3200), Michael Jordan (1700), Paul McCartney (1200}, .J. K. Rowling (1000}, David Coppereld (1000), and Jerry Seinfeld (950) E_|- . Find the range, variance, and standard deviation for the sample data. What do the results tell us about the population of all celebiities? 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 dol ars squared. (Round to the nearest integer as needed.) The standard deviation is $ million. (Round to the nearest integer as needed.) Whatdo 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 not at all typical for all celebrities. O C. Because the statistics are calculated from the data, the measures of variation cannot tell us about other celebrities. O D. Because the data are from celebrities with the highest net worths, the measures of variation are typical for all celebrities. Based on the nature of the amounts, what can be inferred about their preCision'? Because all of the amounts end With V it appears that they are rounded to the nearest 1' so it would make sense to round the results to the nearest 1' In a Leichtman Research Group survey of 1100 TV households, 75% of them had at least one Internet-connected TV device (for example, Smart TV, standalone streaming device, connected video game console). A marketing executive wants to convey high penetration of Internet-connected TV devices, so he makes the claim that the percentage of all homes with at least one Internet-connected TV device is equal to 79% Test that claim using a 0.05 significance level. Use the P-value method. Use the normal distribution as an approximation to the binomial distribution. Let p denote the population proportion of all homes with at least one Internet-connected TV device. Identify the null and alternative hypotheses. Ho: P Hy : P (Type integers or decimals. Do not round.) Identify the test statistic. z= (Round to two decimal places as needed.) Identify the P-value. P-value = (Round to three decimal places as needed.) State the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. the null hypothesis. There sufficient evidence to the claim that the percentage of all homes with at least one Internet-connected TV device is equal to 79%.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: 26, 32, 46, 40, 27, 29. Use a 0.025 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