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In a study of cell phone usage and brain hemisphen dominance, an Internet survey was e-mailed to 6957 subjects randomly selected from an online group involved with ears. There were 1319 surveys returned Use a 0.01
significance level to test the claim that the return rate is less than 20% Use the P-value method and use the normal distnbution as an approximation to the binomial distrbution
Identify the null hypothesis and alternative hypothesis
Question
Answer:
The null hypothesis is [tex]H_0: p \geq 0.2[/tex]
The alternative hypothesis is [tex]H_1: p < 0.2[/tex]
The p-value of the test is of 0.0154 > 0.01, which means that at the 0.01 significance level, there is not significant evidence that the return rate is less than 20%.
Step-by-step explanation:
Test the claim that the return rate is less than 20%.
At the null hypothesis, we test if the return rate is of at least 20%, that is:
[tex]H_0: p \geq 0.2[/tex]
At the alternative hypothesis, we test if the return rate is less than 20%, that is:
[tex]H_1: p < 0.2[/tex]
Normal distribution as an approximation to the binomial distribution to find the Z-score.
Binomial probability distribution
Probability of exactly x successes on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X – \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
0.2 is tested at the null hypothesis. Sample of 6957.
This means that [tex]p = 0.2, n = 6957[/tex]. So
[tex]\mu = E(X) = np = 6957*0.2 = 1391.4[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{6957*0.2*0.8} = 33.36[/tex]
1319 surveys returned
This means that, due to continuity correction, [tex]X = 1319 + 0.5 = 1319.5[/tex]. So
[tex]Z = \frac{X – \mu}{\sigma}[/tex]
[tex]Z = \frac{1319.5 – 1391.4}{33.36}[/tex]
[tex]Z = -2.16[/tex]
P-value of the test:
The p-value of the test is the probability of 1319 or less people returning the survey, which is the p-value of Z = -2.16.
Looking at the z-table, Z = -2.16 has a p-value of 0.0154.
The p-value of the test is of 0.0154 > 0.01, which means that at the 0.01 significance level, there is not significant evidence that the return rate is less than 20%.