Ask AI on The Internet
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?
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.
A new large language model has been made available since this question was asked.
To generate an updated answer from the new model, click the button below.
Think your answer is better than the AI's? Post it below.
Question Tags
If you want your question answered by an AI, click here.
Post your own comment: