Suppose that a chewing gum company claims that $92 \%$ of dentists recommend their chewing gum. Oliver wants to test if this is true. He surveys 132 randomly selected dentists.
1. Assuming the chewing gum company's claim is true, what is the probability that fewer than $86 \%$ of the dentists surveyed recommend the chewing gum? To answer this question, complete the following steps:
* Let the random variable $\widehat{P}$ be the proportion of 132 randomly sampled dentists who recommend the chewing gum. How is $\widehat{P}$ distributed?
$\widehat{P} \sim N(\square, \square)$
* List the $z$-scores needed to calculate the result. If there is more than one $z$-score, separate the values with a comma.
$\square$
* Assuming the chewing gum company's claim is true, the probability that fewer than $86 \%$ of the dentists surveyed recommend the chewing gum is $\square$ .
2. If Oliver does find that fewer than $86 \%$ of the dentists surveyed recommend the chewing gum, would you doubt the chewing gum company's claim? Why or why not? (Consider this question on your own. You do not have to provide an answer here.)
Answer (1)
The sample proportion P follows a normal distribution: P ∼ N ( 0.92 , 132 0.92 ( 0.08 ) ) .
Calculate the z-score: z = 132 0.92 ( 0.08 ) 0.86 − 0.92 ≈ − 2.54
Find the probability using the z-score: P ( Z < − 2.54 ) ≈ 0.0055
The probability that fewer than 86% of dentists recommend the gum is approximately: 0.0055 .
Explanation
Understand the problem and provided data We are given that a chewing gum company claims that 92% of dentists recommend their chewing gum. Oliver surveys 132 dentists and we want to find the probability that fewer than 86% of the dentists surveyed recommend the chewing gum, assuming the company's claim is true. We will use the normal approximation to the sampling distribution of the sample proportion.
Determine the distribution of the sample proportion The sampling distribution of the sample proportion P is approximately normal with mean p = 0.92 and standard deviation n p ( 1 − p ) , where n = 132 . Thus, we have P ∼ N ( 0.92 , 132 0.92 ( 1 − 0.92 ) ) We calculate the standard deviation: 132 0.92 ( 0.08 ) ≈ 0.0236
Calculate the z-score Next, we calculate the z-score for a sample proportion of 86%: z = n p ( 1 − p ) p − p = 132 0.92 ( 0.08 ) 0.86 − 0.92 = 0.0236 − 0.06 ≈ − 2.54
Calculate the probability Finally, we find the probability that fewer than 86% of the dentists surveyed recommend the chewing gum. This is equivalent to finding P ( P < 0.86 ) , which is the same as finding P ( Z < − 2.54 ) . Using a standard normal table or calculator, we find that P ( Z < − 2.54 ) ≈ 0.0055 Thus, the probability that fewer than 86% of the dentists surveyed recommend the chewing gum is approximately 0.0055.
Examples
This type of probability calculation is useful in various fields such as marketing, political science, and quality control. For instance, a marketing team might want to test if a certain percentage of customers prefer their product. By surveying a sample of customers, they can calculate the probability of observing a sample proportion lower than a certain value, assuming their initial claim is true. If this probability is very low, it might suggest that their initial claim is incorrect. This helps in making informed decisions based on statistical evidence.
