Scrabble In the game of Scrabble, each player begins by drawing 7 tiles from a bag containing 100 tiles. There are 42 vowels, 56 consonants, and 2 blank tiles in the bag. Cait chooses an SRS of 7 tiles. Let p^ be the proportion of vowels in her sample.
(a) Is the 10% condition met in this case? Justify your answer.
(b) Is the Large Counts condition met in this case? Justify your answer.
The Correct Answer and Explanation is :
In the context of Cait drawing 7 tiles from a Scrabble bag containing 100 tiles (42 vowels, 56 consonants, and 2 blanks), we examine two statistical conditions: the 10% condition and the Large Counts condition.
(a) 10% Condition:
The 10% condition asserts that when sampling without replacement, the sample size should not exceed 10% of the population to maintain the assumption of independence among observations. This guideline ensures that the removal of each sampled item does not significantly alter the probabilities of subsequent selections.
In this scenario, Cait’s sample size is 7 tiles, and the total population is 100 tiles. Calculating the percentage:
[
\frac{7}{100} \times 100\% = 7\%
]
Since 7% is less than the 10% threshold, the 10% condition is satisfied. This implies that the assumption of independence among the sampled tiles is reasonable, allowing for the use of certain probabilistic models that rely on independent events.
(b) Large Counts Condition:
The Large Counts condition, also known as the Success/Failure condition, is used to determine if the sampling distribution of a sample proportion can be approximated by a normal distribution. This condition is met when both the expected number of successes (np) and failures (n(1-p)) are at least 10.
Here, we define a “success” as drawing a vowel. The probability of drawing a vowel (p) is:
[
p = \frac{\text{Number of vowels}}{\text{Total number of tiles}} = \frac{42}{100} = 0.42
]
The expected number of vowels in Cait’s sample (np) is:
[
np = 7 \times 0.42 = 2.94
]
The expected number of non-vowels (failures) is:
[
n(1-p) = 7 \times (1 – 0.42) = 7 \times 0.58 = 4.06
]
Both expected counts, 2.94 (vowels) and 4.06 (non-vowels), are less than 10. Therefore, the Large Counts condition is not satisfied. This indicates that the sampling distribution of the proportion of vowels in Cait’s sample may not be well-approximated by a normal distribution. Consequently, inferential methods that rely on normality assumptions may not be appropriate in this context.
In summary, while the 10% condition is met, ensuring the independence of selections, the Large Counts condition is not satisfied, suggesting caution when applying normal approximation methods to the proportion of vowels in Cait’s sample.