Question

1.5.25 Suppose two brothers named Mario and Luigi like to compete by playing a certain video game. Mario thinks he is better at this game than Luigi and sets out to prove it by keeping track of who wins. After playing the game 30 times, Mario won 18 of them (or 60%). Mario then declares that this proves he is obviously the better player. Luigi, who just finished Chapter 1 in his statistics class, realizes that perhaps he and his brother are evenly matched and just by chance Mario won 60% of the last 30 games. Luigi is going to test this by running a test of significance XYour answer is incorrect. Which of the following describes what the parameter is that Luigi should be testing? the long-run proportion of times that Mario loses the long-run number of times that Mario wins the long-run number of times that Mario loses the long-run proportion of times that Mario wins eTextbook and Media

please give answer in a,b,c,d format and give calculation clearly.

Answer 1

The total number of times the game is played is 30. The number of times Mario won the game is 18. that is = = = 0.6 proportion. 30 5 1. Since the Luigi is interested in finding whether Mario is better at the game than him, and Luigi has the data about the proportion win of each player, therefore option (a), (b), and (c) are incorrect. Hence, option (d) is correct which states that the long run proportion of times that Mario wins describes the parameter that Luigi should be testing. Since Luigi likes two-sided tests because that will make it harder to get strong evidence that his brother is better, therefore option (c) is incorrect because it is a one-sided test. Option (b) is incorrect because the null hypothesis in this option should be alternative hypothesis. The required null and alternative hypotheses are as follows: Null: The long run proportion of times Mario wins is 50%. Alt: The long run proportion of times Mario wins is different than 50%. This Null and Alternative hypothesis is described by option (a). Hence, option (a) is correct.

The probability that Mario wins is p =0.6. The null and alternative hypotheses are as follows: Null: H :p=0.5 Alt: H : p=0.5 The Z-statistic is given as follows: Z__ - Po Po (1-P) Calculate the Z-statistic and P-value using MINITAB as follows: STEP 1: Click Stat > Basic Statistics > 1 Proportion. STEP 2: Choose Summarized data. STEP 3: Enter 18 in Number of events, and 30 in Number of trials. STEP 4: Check the box of Perform hypothesized test and enter 0.5 in Hypothesized proportion. STEP 5: Go to Options, enter 95 in Confidence Interval, choose not equals to in Alternative hypothesis, and select Normal approximation in Method. STEP 6: Click OK in both the dialog boxes. The MINITAB output obtained is as follows: Test and Cl for One Proportion Method p: event proportion Normal approximation method is used for this analysis. Descriptive Statistics N Event Sample p 9 5% CI for p 30 18 0.600000 (0.424695, 0.775305) Test Horp = 0.5 Hip 0.5 Null hypothesis Alternative hypothesis Z-Value P-value 1.10 0.273 Hence, the P-value is 0.2730- The rejection rule: If P-value is less than the significance level, then reject the null hypothesis. Since P-value = 0.2730 is greater than significance of 0.05, therefore null hypothesis is not rejected, and it can be concluded that the long run proportion of times Mario wins is 50%, that is Mario and Luigi are evenly matched, that is, both are equally better in winning the game. Hence, we have little to no evidence that long-run proportion of times Mario wins is not 50%, that is, option (d) is correct.

By hit and trial method, that is using different values of in Number of events, and Number of trials, from the options provided, in MINITAB maintaining 60% win of Mario, the number of games needed to played is obtained. For Number of trials = 40 and Number of events = 24, the MINITAB output is as follows: Test and Cl for One Proportion Method p: event proportion Normal approximation method is used for this analysis. Descriptive Statistics N Event Samplep 95% CI for p 40 24 0.600000 (0.448182, 0.751818) Test Null hypothesis Ho: p = 0.5 Alternative hypothesis H P = 0.5 Z-Value P-value 1.26 0.206 The P-value this obtained is 0.206, which is greater than the significance level of 0.05. Therefore, option (d) is incorrect. For Number of trials = 50 and Number of events = 30, the MINITAB output is as follows: Test and Cl for One Proportion Method p: event proportion Normal approximation method is used for this analysis. Descriptive Statistics N Event Samplep 95% CI for p 50 30 0.600000 (0.464210, 0.735790) Test Null hypothesis Alternative hypothesis Z-Value P-value 1.41 0.157 Hop = 0.5 H:p0.5 The P-value this obtained is 0.157, which is greater than the significance level of 0.05. Therefore, option (c) is incorrect.

For Number of trials = 80 and Number of events = 48, the MINITAB output is as follows: Test and Cl for One Proportion Method p: event proportion Normal approximation method is used for this analysis. Descriptive Statistics N Event Sample p 9 5% CI for p 80 48 0.600000 (0.492648, 0.707352) Test Null hypothesis Ho: p = 0.5 Alternative hypothesis Hip + 0.5 Z-Value P-value 1.790.074 The P-value this obtained is 0.074, which is greater than the significance level of 0.05. Therefore, option (a) is incorrect.

For Number of trials = 100 and Number of events = 60, the MINITAB output is as follows: Test and Cl for One Proportion Method p: event proportion Normal approximation method is used for this analysis. Descriptive Statistics N Event Sample p 9 5% CI for p 100 60 0.600000 (0.503982, 0.696018) Test Null hypothesis Alternative hypothesis Z-Value P.Value 2.00 0.046 He: p = 0.5 H.: p=0.5 The P-value this obtained is 0.046, which is less than the significance level of 0.05. Therefore, at least 100 games are to be played until Mario gets a P-value less than 0.05. Hence, option (b) is correct.