1. The scores of students on the ACT college entrance examination in a recent year had a normal distribution with a mean of 18.6 and a standard deviation of 5.9. A simple random sample of 60 students who took the exam is selected for study: a) What is the shape, mean(expected value), and standard deviation of the sampling distribution of the sample mean score for samples of size 60? b) What is the probability that the sample mean is 20 or higher? c) What is the probability that the sample mean falls within 2 points of the population mean? d) What value does the sample mean have be in order for it to be in the top 1% of the sampling distribution? 2. Last year, a national opinion poll found that 43% of all Americans agree that parents should be given vouchers good for education at any public or private school of their choice. Assume that in fact the population proportion is 0.43. A random sample of 350 is to be selected and asked the same question. a) What is the shape, mean(expected value), and standard deviation of the sampling distribution of the sample proportion for samples of size 350? b) What is the probability that the sample proportion will fall within 0.03 of the population proportion? 3. Redo part (b) of problem 2, but assume that the sample size is 700. Does this answer make sense when compared to #2, part (b)? Why?
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To answer these questions, we can use the concepts of sampling distributions and the Central Limit Theorem. Let's break down each question step by step.
1) For the ACT scores:
a) The shape of the sampling distribution of the sample mean for samples of size 60 will be approximately normal due to the Central Limit Theorem. The mean of the sampling distribution (expected value) will be the same as the population mean, which is 18.6. The standard deviation of the sampling distribution (standard error) is calculated by dividing the population standard deviation by the square root of the sample size: 5.9 / √60 ≈ 0.761.
b) To find the probability that the sample mean is 20 or higher, we need to standardize the sample mean using the z-score formula: (sample mean - population mean) / standard deviation. In this case, the z-score will be (20 - 18.6) / 0.761 = 1.84. We can then use a standard normal distribution table or a calculator to find the cumulative probability associated with the z-score. The probability P(z > 1.84) represents the probability of getting a sample mean of 20 or higher.
c) To find the probability that the sample mean falls within 2 points of the population mean, we need to find the area under the normal distribution curve between the values of (18.6 - 2) and (18.6 + 2). We can standardize these values using the z-score formula mentioned above and then find the cumulative probability associated with the z-scores P(-2 < z < 2).
d) To find the value of the sample mean that would be in the top 1% of the sampling distribution, we need to find the z-score that corresponds to the top 1% of the standard normal distribution. This z-score can then be used to calculate the sample mean using the formula: sample mean = z-score * standard deviation + population mean.
2) For the opinion poll:
a) The shape of the sampling distribution of the sample proportion for samples of size 350 will be approximately normal due to the Central Limit Theorem. The mean (expected value) of the sampling distribution will be the same as the population proportion, which is 0.43. The standard deviation of the sampling distribution (standard error) is calculated as the square root of (population proportion * (1 - population proportion)) divided by the square root of the sample size: √[(0.43 * (1 - 0.43)) / 350].
b) To find the probability that the sample proportion will fall within 0.03 of the population proportion, we need to find the z-scores for the lower and upper bounds of the range: (population proportion - 0.03) and (population proportion + 0.03). We can then use these z-scores to find the cumulative probability P(z₁ < z < z₂), where z₁ and z₂ are the standardized z-scores of the lower and upper bounds.
3) Redoing part (b) with a sample size of 700:
The answer for part (b) will make sense when compared to the previous question because as the sample size increases, the standard deviation of the sampling distribution decreases. This means that the sampling distribution becomes narrower, resulting in a smaller range of possible sample proportions. Therefore, the probability of the sample proportion falling within 0.03 of the population proportion will increase when compared to the sample size of 350, indicating a higher level of precision in the estimates.