A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. Of the 29tires surveyed, the mean lifespan was 46,700 miles with a standard deviation of 9,800 miles. Using alpha = 0.05, is the data highly consistent with the claim? Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.) 1.State the distribution to use for the test. (Round your answers to two decimal places.) 2.What is the test statistic? (If using the z distribution round your answers to two decimal places, and if using the t distribution round your answers to three decimal places.) z= ? 3.What is the p-value? (Round your answer to four decimal places.) 4. Construct a 95% confidence interval for the true mean. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your lower and upper bounds to the nearest whole number.)
A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. Of the 29tires surveyed, the
Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is
1.State the distribution to use for the test. (Round your answers to two decimal places.)
2.What is the test statistic? (If using the z distribution round your answers to two decimal places, and if using the t distribution round your answers to three decimal places.)
z= ?
3.What is the p-value? (Round your answer to four decimal places.)
4. Construct a 95% confidence interval for the true mean. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your lower and upper bounds to the nearest whole number.)
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