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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 \$405 per year, with a standard deviation of \$30 per year. What is the z-value rounded to the nearest hundredth? Is there enough evidence to reject the claim?

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To calculate the z-value, we will use the formula: z = (sample mean - population mean) / (standard deviation / sqrt(sample size)) Population mean = the claimed average savings of \$470 Sample mean = the average savings in the random sample of 30 customers, which is \$405 Standard deviation = \$30 Sample size = 30 z = (\$405 - \$470) / (\$30 / sqrt(30)) z = (-\$65) / (\$30 / sqrt(30)) z = -\$65 / (\$30 / 5.477) (Here, we've rounded the 5.477 on the denominator just for simplification) z ≈ -65 / 5.48 z ≈ -11.87 The z-value rounded to the nearest hundredth is -11.87. Now, let's determine if there's enough evidence to reject the claim. When considering the criteria for statistical significance, commonly used critical z-values are ±1.96 for a 95% confidence level and ±2.58 for a 99% confidence level. Since the calculated z-value is -11.87, which is far beyond both critical z-values, there is enough evidence to reject the automobile insurance company's claim that their rates for teenage drivers average \$470 less per year than the same coverage from another company.

June 20, 2023, 7:54 p.m.

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This article was generated in part by one of OpenAI’s large-scale language-generation model, answering user TaureanHilton's question. Images were generated with OpenAI's AI model, DALL·E 2. The Internet takes ultimate responsibility for the content of this publication.