A consumer group wants to know if an automobile insurance company with thousands of customers has an average insurance payout for all their customers that is greater than $500 per insurance claim. They know that most customers have zero payouts and a few have substantial payouts. The consumer group collects a random sample of 18 customers and computes a mean payout per claim of $579.80 with a standard deviation of $751.30.
Is it appropriate for the consumer group to perform a hypothesis test for the mean payout of all customers?
A. Yes, it is appropriate because the population standard deviation is unknown.
B. Yes, it is appropriate because the sample size is large enough, so the condition that the sampling distribution of the sample mean be approximately normal is satisfied.
C. No, it is not appropriate because the sample is more than 10 percent of the population, so a condition for independence is not satisfied.
D. No, it is not appropriate because the standard deviation is greater than the mean payout, so the condition that the sampling distribution of the sample mean be approximately normal is not satisfied.
E. No, it is not appropriate because the distribution of the population is skewed and the sample size is not large enough to satisfy the condition that the sampling distribution of the sample mean be approximately normal.

Answer:

E. No, it is not appropriate because the distribution of the population is skewed and the sample size is not large enough to satisfy the condition that the sampling distribution of the sample mean be approximately normal.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

In this question:

Standard deviation larger than the sample mean means that the distribution is skewed.

By the Central Limit Theorem, when the distribution is skewed, normality is assumed for samples sizes of 30 or higher. In this question, the sample is of 18, which is less than 30, so the hypothesis test is not appropriated, and the correct answer is given by option E.

Child Health and Development Studies (CHDS) has been collecting data about expectant mothers in Oakland, CA since 1959. One of the measurements taken by CHDS is the weight increase (in pounds) for expectant mothers in the second trimester. In a fictitious study, suppose that CHDS finds the average weight increase in the second trimester is 14 pounds. Suppose also that, in 2015, a random sample of 45 expectant mothers have mean weight increase of 16.3 pounds in the second trimester, with a standard deviation of 5.9 pounds. A hypothesis test is done to see if there is evidence that weight increase in the second trimester is greater than 14 pounds. Find the p -value for the hypothesis test. The p -value should be rounded to 4 decimal places.

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