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Question: An automobile insurance company claims that its rates for teenage drivers average $470 less per year than the same coverage from another company. In a random sample of 30 customers, the average savings was $435 per year, with a standard deviation of $65 per year. What is the z-value rounded to the nearest hundredth? Is there enough evidence to reject the claim?

To find the z-value, we will use the formula: z = (sample mean - population mean) / (standard deviation / √n) where, sample mean = 435 population mean = 470 standard deviation = 65 n = 30 (sample size) So, z = (435 - 470) / (65 / √30) z = -35 / (65 / 5.48) z = -35 / 11.86 z = -2.95 Rounded to the nearest hundredth, the z-value is -2.95. To determine if there's enough evidence to reject the claim, we need to compare the z-value to the critical value (associated with a desired level of significance). A common significance level is 0.05. For a two-tailed test at a 0.05 significance level, the critical z-values are approximately -1.96 and 1.96. Since our calculated z-value of -2.95 is lower than the critical value of -1.96, we have enough evidence to reject the claim that the automobile insurance company's rates for teenage drivers are $470 less on average per year.

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