For their statistics project, a group of students want to determine the proportion of LMC students who eat fast food frequently.
1 answer:
Answer:
The number of successes is, <em>x</em> = 22.
The number of failures is, (<em>n</em> - <em>x</em>) = 28.
Yes, normality conditions are met.
Step-by-step explanation:
The complete question is:
For their statistics project, a group of students want to determine the proportion of LMC students who eat fast food frequently. They select a random sample of 50 students and find that 22 report eating fast food frequently (more than 3 times a week.)
What is the count of success? Options are: 22,28, 50
What is the count of failures? Options are: 22,28,50
Can the students use this data to calculate a 95% confidence interval? Options are: Yes, normality conditions are met or No, normality conditions are not met
Solution:
Let <em>X</em> denote the number of LMC students who eat fast food frequently.
A random sample of <em>n</em> = 50 students and find that 22 report eating fast food frequently.
The number of successes is, <em>x</em> = 22.
The number of failures is, (<em>n</em> - <em>x</em>) = 28.
The random variable X follows a Binomial distribution with parameters n = 50 and p = 22/50 = 0.44.
But the sample selected is too large and the probability of success is close to 0.50.
So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:
1. np ≥ 10
2. n(1 - p) ≥ 10
Check the conditions as follows:
Thus, a Normal approximation to binomial can be applied.
Yes, normality conditions are met.
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Step-by-step explanation:
Information given
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Null hypothesis:
Alternative hypothesis:
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