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Ede4ka [16]
3 years ago
15

Suppose that a recent poll of American households about car ownership found that for households with a car, 39% owned a sedan, 3

3% owned a van, and 7% owned a sports car. Suppose that three households are selected randomly and with replacement. What is the probability that at least one of the three randomly selected households own a sports car
Mathematics
1 answer:
Rama09 [41]3 years ago
4 0

Answer:

The probability that of the 3 households randomly selected at least 1 owns a sports car is 0.1956.

Step-by-step explanation:

Let <em>X</em> = number of household owns a sports car.

The probability of <em>X</em> is, P (X) = p = 0.07.

Then the random variable <em>X</em> follows a Binomial distribution with <em>n</em> = 3 and <em>p</em> = 0.07.

The probability function of a binomial distribution is:

P(X=x) = {n\choose x}p^{x}[1-p]^{n-x}\\

Compute the probability that of the 3 households randomly selected at least 1 owns a sports car:

P(X\geq 1)=1-P(X

Thus, the probability that of the 3 households randomly selected at least 1 owns a sports car is 0.1956.

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Answer:

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p_v =P(z>0.6)=0.274  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can't conclude that the true mean is not significantly higher than 200 at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=203 represent the mean height for the sample  

\sigma=15 represent the population standard deviation

n=9 sample size  

\mu_o =200 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 200, the system of hypothesis would be:  

Null hypothesis:\mu \leq 200  

Alternative hypothesis:\mu > 20  

If we analyze the size for the sample is < 30 but we  know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

z=\frac{203-200}{\frac{15}{\sqrt{9}}}=0.6    

P-value

Since is a one sided test the p value would be:  

p_v =P(z>0.6)=0.274  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can't conclude that the true mean is not significantly higher than 200 at 5% of signficance.  

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