After voting in the latest round of California elections, a large group of voter
ID: 3252162 • Letter: A
Question
After voting in the latest round of California elections, a large group of voters in 10 California counties are asked to grade the character of the campaigns for four propositions. A 0% indicates deep dissatisfaction, 50% neutral, and 100% utter and blissful satisfaction (we can assume drugs are involved here). The mean for this group is 43% with a standard deviation of 14.6%. Task 1: Individual normal distribution scores Assuming that satisfaction ratings are normally distributed: a. What percentage rated the campaigns at 43% or higher? b. what percentage rated the campaigns above a low 'C' or 70%? c. What percentage rated the campaigns 'minimally passing' (60%) or better? d. What percentage were so disgusted that they rated the campaigns no better than 30%? Task 2: Distribution of sample means Same scenario, but this time that group is a random sample drawn from those 10 counties. The size of the sample is 750 (N). Using our two curve (Chapter 6) 95% confidence limit test**: a. What would we estimate the true mean rating to be from the population from which this sample was randomly drawn? b. What population mean values can we confidentially (95%) reject as possibilities? Use a phrase such as "any value above... or....". c. Might the sample have been randomly drawn from a population outside of that estimated range? Why or why not? d. Can we specify exactly what the true population mean is? Why or why not? e. What would the range of estimates be with a sample twice that size (1500)?Explanation / Answer
Solution:
Task 1:
Population Mean = = 43% = 0.43
Population Standard Deviation = = 14.6 = 0.146
Solution - A:
z-score for (x = 43% = 0.43):
z = (x – ) / = (0.43 – 0.43) / 0.146
z = 0
Percentage of voters rating the campaigns at 43% or higher = P(z 0) = 0.5 = 50%
Hence, percentage of voters rating the campaigns at 43% or higher = 50%
Solution – B:
Z – score for (x = 70% = 0.70):
z = (x – ) / = (0.70 – 0.43)/0.146
z = 1.849
Percentage of voters rating the campaigns above a low ‘C’ or 70% = P(z > 1.849)
From normal distribution table, P(z > 1.849) = 1 – 0.9678 = 0.0322 = 3.22%
Hence, percentage of voters rating the campaigns above a low ‘C’ or 70% = 3.22%
Solution – C:
Z – score for (x = 60% = 0.60):
z = (x – ) / = (0.60 – 0.43) / 0.146
z = 1.164
Percentage of voters rating the campaigns ‘minimally passing’ (60%) or better = P(z 1.164)
From normal distribution table, P(z > 1.164) = 1 – 0.8779 = 0.1221 = 12.21%
Hence, Percentage of voters rating the campaigns ‘minimally passing’ (60%) or better = 12.21%
Solution – d:
Z – score for (x = 30% = 0.30):
z = (x – ) / = (0.30 – 0.43) / 0.146
z = - 0.890
Percentage of voters rating the campaigns no better than 30% = P(z -0.890) = 0.1866 = 18.66%
Hence, Percentage of voters rating the campaigns no better than 30% = 18.66%