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Fynjy0 [20]
3 years ago
11

one town, 64% of adults have health insurance. What is the probability that 10 adults selected at random from the town all have

health insurance?
Mathematics
1 answer:
Katena32 [7]3 years ago
3 0

Answer:

The probability that 10 adults selected at random from the town all have health​ insurance is 0.01153.

Step-by-step explanation:

Consider the provided information.

One town, 64% of adults have health insurance.

Let p = 64% = 0.64

Therefore, q=1-0.64=0.36

We need to find the probability that 10 adults selected at random from the town all have health insurance

Use the formula: P(x)=^nC_r(p)^r(q)^{n-r}

Here, the value of r is 10.

Substitute the respective values in the above formula.

P(x)=^{10}C_{10}(0.64)^{10}(0.36)^{10-10}

P(x)=(0.64)^{10}(0.36)^{0}\\P(x)=(0.64)^{10}\\P(x)\approx0.01153

Hence, the probability that 10 adults selected at random from the town all have health​ insurance is 0.01153.

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Answer:

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If we compare the p value and the significance level given \alpha=0.01 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 conclude that the true mean is not significantly higher than 48.8 at 1% of signficance.  

Step-by-step explanation:

1) Data given and notation  

\bar X=53.4 represent the sample mean

\sigma=12 represent the population standard deviation assumed

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\alpha=0.01 represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

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2) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is higher than 48.8, the system of hypothesis would be:  

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Alternative hypothesis:\mu > 48  

Since we assume that know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

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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".  

3) Calculate the statistic  

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

z=\frac{53.4-48.8}{\frac{12}{\sqrt{36}}}=2.3  

4)P-value  

Since is a right tailed test the p value would be:  

p_v =P(Z>2.3)=1-P(Z  

5) Conclusion  

If we compare the p value and the significance level given \alpha=0.01 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 conclude that the true mean is not significantly higher than 48.8 at 1% of signficance.  

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