Extra Examples for Section 11.1
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1 NAME: Extra Examples for Section 11.1 ere are the results of Jenny s 6 rolls of her ceramic die and the expected counts. Outcome Observed Expected Total 6 6 Problem: Calculate the value of the chi-square statistic. When Jenny rolled her ceramic die 6 times and calculated the chi-square statistic, she got = 3.4. Problem: Using the appropriate degrees of freedom, calculate the P-value. What conclusion can you make about Jenny s die?
2 Landline surveys According to the Census, of all US residents age and older, 19.1% are in their s, 1.5% are in their 3 s, 1.1 % are in their 4 s, 15.5% are in their 5 s, and.8% are 6 and older. The table below shows the age distribution for a sample of US residents age and older. Members of the sample were chosen by randomly dialing landline telephone numbers. Category Count Total 148 Do these data provide convincing evidence that the age distribution of people who answer landline telephone surveys is not the same as the age distribution of all US residents?
3 Birthdays and hockey In his book Outliers, Malcolm Gladwell suggests that a hockey player s birth month has a big influence on his chance to make it to the highest levels of the game. Specifically, since January 1 is the cut-off date for youth leagues in Canada (where many NL players come from), players born in January will be competing against players up to 1 months younger. The older players tend to be bigger, stronger, and more coordinated and hence get more playing time, more coaching, and have a better chance of being successful. To see if this is true, a random sample of 8 National ockey League players from the 9-1 season was selected and their birthdays were recorded. Overall, 3 were born in the first quarter of the year, in the second quarter, 16 in the third quarter, and 1 in the fourth quarter. Do these data provide convincing evidence that the birthdays of NL players are not uniformly distributed throughout the year?
4 Solution: = = 3.4. Solution: Since there are six possible outcomes when rolling her die, the degrees of freedom = 6 1 = 5. Using Table C, the P-value is greater than.5 since the statistic is smaller than the lowest critical value in the df = 5 row. Using technology, cdf(3.4,1,5) =.64. Since the P-value is quite large, we do not have convincing evidence that her die is unfair. owever, this doesn t prove that her die is fair. Landline surveys State: We want to perform a test of the following hypotheses using =.5: : The age distribution of people who answer landline telephone surveys is the same as the age a : The age distribution of people who answer landline telephone surveys is not the same as the age Plan: If conditions are met, we will perform a chi-square goodness-of-fit test. Random The data came from a random sample of US residents who answer landline telephone surveys. Large Sample Size The expected counts are 148(.191) =., 148(.15) = 5.3, 148(.11) = 1.1, 148(.155) = 16.4, 148(.8) = All expected counts are at least 5. Independent Because we are sampling without replacement, there must be at least 1(148) = 1,48 U.S. residents who answer landline telephone surveys. This is reasonable to assume. Do: 141. Test Statistic 48.. P-value Using 5 1 = 4 degrees of freedom, P-value = cdf(48.,1,4). Conclude: Because the P-value is less than =.5, we reject. We have convincing evidence that the age distribution of people who answer landline telephone surveys is not the same as the age
5 Birthdays and hockey State: We want to perform a test of the following hypotheses using =.5: : The birthdays of NL hockey players are equally likely to occur in each quarter of the year. a : The birthdays of NL hockey players are not equally likely to occur in each quarter of the year. Plan: If conditions are met, we will perform a chi-square goodness-of-fit test. Random The data came from a random sample of NL players Large Sample Size If birthdays are equally likely to be in each quarter of the year, then the expected counts are all =. These counts are all at least 5. Independent Because we are sampling without replacement, there must be at least 1(8) = 8 NL hockey players. In the 9-1 season, there were 879 NL players, so this condition is met. Do: 3 Test Statistic 11. P-value Using 4 1 = 3 degrees of freedom, P-value = cdf(11.,1,3)=.11. Conclude: Because the P-value is less than =.5, we reject. We have convincing evidence that the birthdays of NL players are not uniformly distributed throughout the year.
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