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Is it appropriate to use the normal distribution to approximate the sampling distribution of \(\hat{p}\) for the situations described in Exercises \(4-8 ?\) $$n=50, p=.05$$

Short Answer

Expert verified
Answer: No, it is not appropriate to use the normal distribution to approximate the sampling distribution of \(\hat{p}\) for the given situation since the condition np ≥ 10 is not met.

Step by step solution

01

Verify the sample is random

In this exercise, we are not given specific information about the sample selection process, so we will assume it is randomly selected.
02

Check the sample size conditions

We will calculate np and n(1-p) and check if both are greater than or equal to 10. np = \(50 \times 0.05 = 2.5\) n(1-p) = \(50 \times (1 - 0.05) = 50 \times 0.95 = 47.5\) We see that np = 2.5 and n(1-p) = 47.5. While n(1-p) is greater than or equal to 10, np is less than 10.
03

Conclusion

Since the condition np ≥ 10 is not met, it is not appropriate to use the normal distribution to approximate the sampling distribution of \(\hat{p}\) for the given situation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Distribution
The normal distribution is a continuous probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. In the context of sampling distributions, the normal distribution can be used as an approximation under certain conditions.

This is beneficial because the normal distribution has well-known properties, making it easier to calculate probabilities and critical values. However, in order for the normal distribution to provide an accurate approximation of the sampling distribution of a sample proportion \( \hat{p} \), specific conditions must be met. These include the sample being random and the sample size being sufficiently large to ensure the product of the sample size and the population proportion \( n \cdot p \), as well as \( n \cdot (1 - p) \), are both greater than or equal to 10.

If these conditions are not met, the approximation may lead to significant errors. As seen in the exercise, the condition \( np \geq 10 \) was not satisfied, meaning that the use of normal distribution in this situation would not be appropriate.
Binomial Distribution Conditions
The binomial distribution represents the number of successes in a sequence of independent experiments, each asking a yes/no question, and each with its own boolean-valued outcome: success or failure. Several conditions need to be met for a distribution to be classified as binomial:
  • The number of observations or trials is fixed.
  • Each trial must be independent of each other.
  • There are only two possible outcomes for each trial, often labeled as 'success' and 'failure'.
  • The probability of success \( p \) must be the same for each trial.

For the approximation of binomial distribution by the normal distribution, it's required that both \( np \) and \( n(1-p) \) are greater than or equal to 10, ensuring that the binomial distribution is symmetric enough to resemble the normal curve. Where this symmetry is not present, as in the exercise provided, the normal approximation is not suitable, and the exact binomial distribution should be used instead.
Sample Size
Sample size is a critical factor in statistics, as it influences the accuracy and precision of the estimation or hypothesis testing. A larger sample size generally leads to more reliable and robust results because it reduces the sample variability and draws the sampling distribution closer to the normal distribution, known as the Central Limit Theorem.

The choice of sample size also affects the ability of a statistician to detect a significant effect. In cases where the sample size is too small, even a substantial effect may not be statistically significant, simply due to the high variability inherent in small samples. Conversely, a very large sample size might lead to finding statistically significant differences that are not practically meaningful.

Considering the exercise, the sample size of 50 did not meet the threshold for \( np \) to be at least 10. This suggests that increasing the sample size could improve the approximation of the sampling distribution of \( \hat{p} \) to a normal distribution, thus providing more accurate results in the estimation of the population proportion.

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Most popular questions from this chapter

Baby! The weights of 3 -month-old baby girls are normally distributed with a mean of 5.9 kilograms and a standard deviation of \(0.7 .^{6}\) In an inner city pediatric facility, a random sample of 16 -month-old baby girls was selected and their weights were recorded. a. What is the probability that the average weight of the 16 baby girls is more than 6.4 kilograms? b. What is the probability that the average weight of the 16 baby girls is less than 5.0 kilograms? c. If the average weight of the 16 girls at this inner city pediatric facility was 5.0 kilograms, what conclusions might you draw? Explain.

For the binomial experiments described in Exercises \(24-26,\) describe the approximate shape of the sampling distribution of \(\hat{p}\) and calculate its mean and standard deviation (or standard error). Then calculate the probability given in the exercise. A random sample of size \(n=50\) is selected from a binomial distribution with \(p=.7\). Find the probability that the sample proportion \(\hat{p}\) is less than .8 .

The manager of a building supplies company randomly samples incoming lumber to see whether it meets quality specifications. From each shipment, 100 pieces of \(2 \times 4\) lumber are inspected and judged according to whether they are first (acceptable) or second (defective) grade. The proportions of second- grade \(2 \times 4\) s recorded for 30 shipments were as follows: \(\begin{array}{llllllllll}.14 & .21 & .19 & .18 & .23 & .20 & .25 & .19 & .22 & .17 \\ .21 & .15 & .23 & .12 & .19 & .22 & .15 & .26 & .22 & .21 \\ .14 & .20 & .18 & .22 & .21 & .13 & .20 & .23 & .19 & .26\end{array}\) a. Construct a control chart for the proportion of second-grade \(2 \times 4 \mathrm{~s}\) in samples of 100 pieces of lumber. b. Explain how the control chart can be of use to the manager of the building- supplies company.

Suppose that college faculty with the rank of professor at public 2-year institutions earn an average of \(\$ 75,878\) per year \(^{7}\) with a standard deviation of \(\$ 4000\). In an attempt to verify this salary level, a random sample of 60 professors was selected from an appropriate database. a. Describe the sampling distribution of the sample \(\operatorname{mean} \bar{x} .\) b. Within what limits would you expect the sample average to lie, with probability \(.95 ?\) c. Calculate the probability that the sample mean \(\bar{x}\) is greater than \(\$ 78,000 ?\) d. If your random sample actually produced a sample mean of \(\$ 78,000\), would you consider this unusual? What conclusion might you draw?

Is it appropriate to use the normal distribution to approximate the sampling distribution of \(\hat{p}\) for the situations described in Exercises \(4-8 ?\) $$n=75, p=.1$$

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