Sampling Distribution Model Quiz Questions

As the sample size increases, the sampling distribution model for the sample mean becomes more Normal in shape. This is because the Central Limit Theorem states that as the sample size increases, the distribution of sample means approaches a normal distribution regardless of the shape of the population distribution. Additionally, the standard deviation of the sampling distribution gets smaller as the sample size increases. This is because larger sample sizes provide more information and reduce the variability of the sample means, resulting in a smaller standard deviation.

The mean of the proportion of homeowners in this sample who prefer a finished basement is μ = 42%. This means that, on average, 42% of homeowners in the sample prefer a finished basement.

The randomization condition is not satisfied because the students were selected only from those leaving the student union building, not from the whole student body. This means that the sample may not be representative of all students at the college, especially those who are more likely to stay in their rooms and not participate in activities at the student union building, such as those suffering from depression. Therefore, the Normal model may not be used to describe the distribution of the proportion of students in the sample who suffer from depression.

The standard deviation of the proportion of students who may wear contact lenses in this group is 2.52%. This means that the actual proportion of students who wear contact lenses in different samples of 200 students is expected to vary by about 2.52% on average.

The explanation for the given correct answer is that the distribution of incomes in the original population is skewed to the right. In order to use the Normal model to describe the sampling distribution of the mean, the sample size should be large enough for the Central Limit Theorem to apply. However, a sample size of 12 is not considered large enough when the population distribution is skewed. Therefore, the Normal model may not be used in this case.

If the population is strongly skewed to the right and the sample size is small, the sampling distribution model for the sample mean would still be skewed right, with the center at the population mean (μ), and the standard deviation would be σ/sqrt(n), where σ is the standard deviation of the population and n is the sample size. This is because even though the sample mean tends to be less skewed than the population, it still retains some of the skewness, and the standard deviation decreases as the sample size increases.

If a population is strongly skewed to the right, using a large sample rather than a small one will not affect the shape of the distribution of the sample data (I). However, using a large sample will result in a sampling model of the sample means that is closer to normal (II). This is because the Central Limit Theorem states that as sample size increases, the distribution of sample means approaches a normal distribution, regardless of the shape of the population distribution. The variability of the sample means (III) will be smaller with a larger sample size, as it reduces the effect of outliers and extreme values. Therefore, the correct answer is II only.

The probability that the shipment will be accepted anyway can be calculated using the concept of binomial distribution. In this case, we have a sample size of 125, with a probability of 9% (0.09) for an orange to be unsatisfactory. We want to find the probability that the number of unsatisfactory oranges in the sample is less than or equal to 8% (0.08) of the sample. By calculating this probability using the binomial distribution formula, we find that the probability is 0.3483.

If we use a large sample rather than a small one, the distribution of our sample data will be more representative of the population, which means it will be less skewed to the left. Therefore, statement I is true. However, the sampling model of the sample means will not be affected by the size of the sample, so statement II is false. The variability of the sample means will actually be smaller with a larger sample, as it reduces the impact of outliers and random errors, so statement III is false. Therefore, the correct answer is I only.

The probability that the next 100 customers will spend an average of at least $50 on dinner is 0.9868. This means that there is a high likelihood that the average spending of the next 100 customers will be at least $50.