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2 years ago

15

Assuming that the sample mean carapace length is greater than 3.39 inches, what is the probability that the sample mean carapace

length is more than 4.03 inches

1 answer:

joja [24]2 years ago

4 0

Answer:

The answer is "".

Step-by-step explanation:

Please find the complete question in the attached file.

We select a sample size n from the confidence interval with the mean $\mu$ and default $\sigma$ , then the mean take seriously given as the straight line with a z score given by the confidence interval

$\mu=3.87\\\\\sigma=2.01\\\\n=110\\\\$

Using formula:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

The probability that perhaps the mean shells length of the sample is over 4.03 pounds is

$P(X>4.03)=P(z>\frac{4.03-3.87}{\frac{2.01}{\sqrt{110}}})=P(z>0.8349)$

Now, we utilize z to get the likelihood, and we use the Excel function for a more exact distribution

$=\textup{NORM.S.DIST(0.8349,TRUE)}\\\\P(z$

the required probability: $P(z>0.8349)=1-P(z$

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Answer and Step-by-step explanation:

This is a complete question

Trials in an experiment with a polygraph include 97 results that include 23 cases of wrong results and 74 cases of correct results. Use a 0.01 significance level to test the claim that such polygraph results are correct less than 80% of the time. Identify the nullhypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution.

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Now Test statistic = z

$= \hat p - P0 / [\sqrtP0 \times (1 - P0 ) / n]$

$= 0.7629 - 0.80 / [\sqrt(0.80 \times 0.20) / 97]$

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