Probability Distribution Issues - Speech or Presentation Example

Answer The given frequency table ifies data into non-overlapping intervals. Knowing this, the probability of each interval can easily be taken. First, the total is determined as follows: The probability of each interval is then simply the frequency divided by the total. The probabilities are tabulated below. Waiting time (minutes) 0-3 4-7 8-11 12-15 16-19 20-23 24-27 Probability 0.2692 0.1731 0.2115 0.1154 0.1346 0.0577 0.0385 The probability of a random customer waiting at least 12 minutes is therefore: Similarly, the probability of a customer waiting between 8 and 15 minutes is: Combining both conditions is not simply a matter of adding the two probabilities as the two sets are not mutually exclusive. The condition occurs in both conditions. Taking this into consideration, the total probability of a customer waiting at least 12 minutes or between 8 and 15 minutes is: Answer 2 While there may be 22 possible digits, each digit can be taken as a binary decision. Particularly, each digit can be tested on whether it matches with the drawn number or not. The resulting problem can then be treated as a binomial probability distribution. The probability of getting the right digit and its corresponding complement is given as: A winning ticket is only possible when all six digits are matched up. Therefore, the probability equation can be constructed as follows: The probability of winning the lottery is then around nine in a billion. It should be noted that the resulting equation actually simplifies to a multiplication rule. However, the binomial distribution may be used to model more complex lottery options. Answer 3 In the construction of a 5-digit number, the order in which is numbers are drawn is clearly important. Also, since repetition of digits is not allowed, a single digit chosen is removed from the pool of possible succeeding digits. These conditions indicate that the total number of digits can simply be taken using a permutation. Answer 4 The given table can only be classified as a probability distribution if its individual probabilities are between zero and one and the total probability is equal to one. In the given table, the probability at X = 2 is greater than one. Since no negative probabilities are present, the total probability is also greater than one. As both requirements are not met, the table cannot be a probability distribution. Answer 5 Inspecting the individual values, they all fall within the range of zero to one. This means that requirement for individual probabilities are met. The total probability can then be taken: Since the probabilities do not total to one, the given data cannot be labeled as a probability distribution. Answer 6 To determine the probability of a credit rating of 60 and below, the corresponding z-score is determined first. Inspecting the negative z-score distribution table, this corresponds to a probability of 0.0026. Answer 7 Given the sample data, the proportion for the sample population and its complement are given as: At 99% confidence interval the appropriate z-score for a two-tailed analysis is: The error of estimate can then be determined: The estimated population proportion is then: or Answer 8 The sample data is relatively sparse (n < 30) which warrants the use of the t-distribution instead of z-scores. To verify whether or not the population mean lifetimes approximate that of the advertized mean, it is necessary to obtain the sample mean and standard deviation. To aid in this, the following table is constructed: 995 590 510 539 739 10861 917 571 555 916 728 664 693 708 887 849 990025 348100 260100 290521 546121 8208841 840889 326041 308025 839056 529984 440896 480249 501264 786769 720801 The mean and standard deviation can then be computed as follows: Based on these parameters, it is then possible to determine whether the advertized mean differs from the actual mean. Hypotheses: H0: The advertized mean is the same as the actual mean. H1: The advertized mean differs from the actual mean. Degrees of Freedom: From the t-distribution table, the corresponding critical value for two tails at 90% confidence is 1.761. Decision Rule: Reject H0 if 1.761 Calculations: Conclusion: Since 1.761, H0 is rejected. At 90% confidence, there is sufficient evidence to support that the advertized and actual mean life times differ. Answer 9 At a sample size of 23, it is appropriate to use the t-distribution in place of z-scores. Since the relevant sample statistics have been provided, the claim can then be directly tested. Hypotheses: H0: The mean lifetime of the engine is approximately 220,000 miles. H1: The mean lifetime of the engine is greater than 220,000 miles. Degrees of Freedom: The critical value for a one-tailed analysis at 99% confidence is given as 2.508. Decision Rule: Reject H0 if 2.508 Calculations: Conclusion: Since 2.508, H0 is rejected. At 99% confidence, there is sufficient evidence to support that the mean lifetime of a Dodge Hemi engine exceeds 220,000 miles. Answer 10 To determine whether a relationship exists between the two variables, the given data may be fitted onto a function. The simplest approach is to use a linear model for approximating the relation. To aid in this, a column representation may be constructed. It should be noted that the independent variable based on the description is the cost of the product. 9 81 85 7225 765 2 4 52 2704 104 3 9 55 3025 165 4 16 68 4624 272 2 4 67 4489 134 5 25 86 7396 430 9 81 83 6889 747 10 100 73 5329 730 44 320 569 41681 3347 The given totals can then be used to construct the model: Hypotheses: H0: The advertising cost is not linearly related to the quantity sold. H1: The advertising cost is linearly related to the quantity sold. The critical values for n = 8 are 0.707 and 0.834 at α = 0.05 and 0.01 respectively. Decision Rule: Reject H0 if 0.707 (95% confidence) or 0.834 (99% confidence) Conclusion: Since 0.707, reject H0. At 95% confidence, there exists a linear relationship between the advertising cost and the quantity sold. Answer 11 An assumed frequency distribution is assumed in this case which allows the use of the Goodness of Fit test. The computations for the chi square statistic can then be determined as shown in the following table: Day Monday Tuesday Wednesday Thursday Friday 37 15 12 23 43 26 26 26 26 26 11 -11 -14 -3 17 121 121 196 9 289 4.6538 4.6538 7.5385 0.3462 11.1154 Hypotheses: H0: Absences occur at the same frequency for different days of the week. H1: Absences occur at different frequencies for different days of the week. Degrees of Freedom: At 95% confidence and 4 degrees of freedom, the critical value for the chi-square statistic is 9.488. Decision Rule: Reject H0 if 9.488 Conclusion: Since 9.488, reject H0. At 95% confidence level, the number of absences differs depending on the day of the week. Answer 12 Data is a very important aspect of any experiment. However, for any given experiment to be successful, there are actually two aspects of data that become necessary. The more apparent one is data analysis. Proper analysis of data can reveal certain characteristics and trends which may be more relevant than the raw data itself. For instance, the use of regression analysis in a data set can reveal a trend between variables which may be used to interpolate or extrapolate points. Data can be analyzed to predict the relationship between a sample set and the actual population. By manipulating raw data, a wide range of information can become available. A more subtle aspect of data analysis, however, is that of data collection. Data itself is not simply data. It is the result of a carefully thought out process of obtaining information. Faulty data collection can mean that any information obtained from the data may likewise be unreliable. For instance, the addition of biased criteria in sample selection would mean that while trends may exist in the data, the actual situation may be much more different. In regression analysis, hidden variables may contribute to a trend but remain unaccounted for. In reality, experimental design is highly critical to the success of the experiment. Answer 13 To determine whether the two variables are independent, a chi-square Test of Independence may be used. The following contingency table presents the relevant calculations for the chi-square statistic: Yes No Undecided Employed 30 (50)(50)/105 = 23.81 15 (50)(40)/105 = 19.05 5 (50)(15)/105 = 7.14 50 Unemployed 20 (55)(50)/105 = 26.19 25 (55)(40)/105 = 20.95 10 (55)(15)/105 = 7.86 55 50 40 15 105 Hypotheses: H0: The response to the questionnaire is independent on the employment status. H1: The response to the questionnaire is dependent on the employment status. Degrees of Freedom: At 90% confidence and 2 degrees of freedom, the critical value for the chi-square statistic is 4.605. Decision Rule: Reject H0 if 4.605 Conclusion: Since 4.605, reject H0. At 90% confidence level, there is sufficient evidence to suggest that the response to the questionnaire is dependent on the employment status of the respondent. Answer 14 In this problem, the independence of three sample groups is to be tested. Since the samples do not fall into discrete categories, the chi-square distribution cannot be used to analyze the data. Normal distribution comparison through the t-test is also not possible due to the number of groups to be compared. An alternative is a generalized form of the t-test known as the one-way analysis of variance. The following table can be constructed to analyze the data: Brand 1 Brand 2 Brand 3 260 181 238 218 240 257 184 162 241 219 218 213 881 801 949 220.25 200.25 237.25 The overall mean is: The mean square value between groups is computed as follows: The table is then reconstructed by centering each group about its respective mean. This is done by subtracting the mean from each value: Brand 1 Brand 2 Brand 3 39.75 -19.25 0.75 -2.25 39.75 19.75 -36.25 -38.25 3.75 -1.25 17.75 -24.25 Again, the mean square value is determined. However, in this particular case, the mean value is obtained within groups. Squaring each element in the table yields the following: Brand 1 Brand 2 Brand 3 1580.06 370.56 0.56 5.06 1580.06 390.06 1314.06 1463.06 14.06 1.56 315.06 588.06 The computations following this are shown below: The F-statistic used in this test can then be determined using the ratio of the mean squares: From an F-distribution table, the critical value with the 2 and 9 degrees of freedom using a 0.05 significance level is 4.26. Hypotheses: H0: The pens of the three brands have the same mean lifetimes. H1: The lifetimes of the pens depend on the particular brand. Decision Rule: Reject H0 if 4.26 Conclusion: Since 4.26, H0 cannot be rejected. At 95% confidence level, there is insufficient evidence to support that the lifetimes of pens of the three brands differ. Answer 15 The predicted score for the student is 80.6. Answer 16 Before attempting to predict the value for y, it is necessary to determine whether the given equation is capable of modeling the data. For a sample size of 4, the critical value at 5% significance level is at 0.950. Since the correlation coefficient is lower than the given critical value, it can be concluded that the given data set cannot effectively be modeled through linear regression. Consequently, a best predicted value cannot be given using the supplied information. Answer 17 a. H0: The Penske team’s engine has approximately the same fuel efficiency as the Honda engine. H1: The Penske team’s engine has greater fuel efficiency compared to the Honda engine. b. H0: Health care coverage is independent of gender. H1: Health care coverage is dependent on gender. c. H0: The average August temperature in Norman is not more than the average August temperature in New York City. H1: The average August temperature in Norman is more than the average August temperature in New York City. d. H0: The student’s athletic ability does not affect his score on a math CLEP test. H1: The student’s athletic ability is related to his score on a math CLEP test. Answer 18 The given problem requires the probability of the right tail of the normal distribution. This can be determined by taking the left tailed probability from the z-score tables. The selected point must first be converted into its corresponding z-score: The corresponding probability from the z-score table is 0.8413. The probability of a randomly selected teach earning more than $525 a week is then given as: Answer 19 Since the problem supplies the population mean and standard deviation, the sample statistics can be obtained using the central limit theorem. The corresponding z-score is then determined as follows: From the negative z-score tables, the probability of the rebuild time being less than 7.6 hours is then given as 0.0025. Answer 20 The statement “fail to reject the null hypothesis” is more appropriate due to the nature of the normal probability distribution. In such a distribution, probabilities are concentrated on a central lobe and rapidly drop as the difference from the mean increases. The null hypothesis states that the actual mean falls within the accepted portion of the curve. On the other hand, the alternative hypothesis places the actual mean on the extreme ends of the curve. Since these extremities have very low probabilities, the reject of the null hypothesis confidently places the actual mean at a very narrow portion of the curve. Failing to reject the null hypothesis, on the other hand, means that the actual mean falls within the central limits. However, since this central area is large (particularly for high confidence levels), the actual mean may still fall near one of the extreme ends but within the accepted limits. This means that a trend exists but there is just insufficient evidence to support that it is indeed significant. The null hypothesis cannot be accepted as the actual mean cannot surely be placed near the population mean. Read More