A shareholders' group, in lodging a protest, claimed that the mean tenure for a chief executive office (CEO) was at least seven
1 answer:
Complete question is;
A shareholders’ group, in lodging a protest, claimed that the mean tenure for a chief executive office (CEO) was at least seven years. A survey of companies reported in The Wall Street Journal found a sample mean tenure of x¯ = 7.27 years for CEOs with a standard deviation of s = 6.38 years (The Wall Street Journal, January 2, 2007).
a. Formulate hypotheses that can be used to challenge the validity of the claim made by
the shareholders’ group.
b. Assume 85 companies were included in the sample. What is the p-value for your hypothesis
test?
Answer:
A) Null hypothesis: H0: μ ≥ 7
Alternative hypothesis: μ < 7
B) p-value = 0.3488
Step-by-step explanation:
A) We are told that the shareholders group claimed that the mean tenure for a chief executive office (CEO) was at least seven years.
Thus, defining the hypothesis, we have;
Null hypothesis: H0: μ ≥ 7
Alternative hypothesis: μ < 7
B) Sample mean: x¯ = 7.27 years standard deviation: s = 6.38 years
Sample size = 85
Formula for the test static is;
t = (x¯ - μ)/(s/√n)
t = (7.27 - 7)/(6.38/√85)
t = 0.39
From online p-value from t score calculator attached using: t = 0.39, DF= 85 - 1 = 84, significance value = 0.05, one tail, we have;
p-value = 0.3488
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Answer:
95.64% probability that under 30% of the dogs are emotional support trained
Step-by-step explanation:
For each dog, there are only two possible outcomes. Either they have completed emotional support training, or they have not. So we use the binomial probability distribution to solve this problem.
However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.
Binomial probability distribution
Probability of exactly x sucesses 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:
The standard deviation of the binomial distribution is:
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
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 pvalue, 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 ,
.
In this problem, we have that:
What is the probability that under 30% of the dogs are emotional support trained?
30% of 100 is 0.3*100 = 30
So this is the pvalue of Z when X = 30.
has a pvalue of 0.9564.
So there is a 95.64% probability that under 30% of the dogs are emotional support trained