A consumer products testing group is evaluating two competing brands of tires, Brand 1 and Brand 2. Tread wear can vary considerably depending on the type of car, and the group is trying to eliminate this effect by installing the two brands on the same random sample of 10 cars. In particular, each car has one tire of each brand on its front wheels, with half of the cars chosen at random to have Brand 1 on the left front wheel, and the rest to have Brand 2 there. After all of the cars are driven over the standard test course for 20,000 miles, the amount of tread wear (in inches) is recorded, as shown in Table 1. Car Brand 1 Brand 2 Difference (Brand 1 - Brand 2) 1 0.362 0.235 0.127 2 0.343 0.368 -0.025 3 0.203 0.230 -0.027 4 0.335 0.247 0.088 5 0.207 0.221 -0.014 6 0.344 0.332 0.012 7 0.354 0.285 0.069 8 0.385 0.382 0.003 9 0.273 0.236 0.037 10 0.376 0.328 0.048 Table 1 Based on these data, can the consumer group conclude, at the 0.05 level of significance, that the mean tread wears of the brands differ? Answer this question by performing a hypothesis test regarding μd (which is μ with a letter "d" subscript), the population mean difference in tread wear for the two brands of tires. Assume that this population of differences (Brand 1 minus Brand 2) is normally distributed. Perform a two-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.) The null hypothesis: H0: _____ The alternative hypothesis: H1: ______ The type of test statistic: (Choose one) Z, t, Chi- square, and F The value of the test statistic: (Round to at least three decimal places.) The two critical values at the 0.05 level of significance: (Round to at least three decimal places.) _____ and ____ At the 0.05 level, can the consumer group conclude that the mean tread wears of the brands differ? Yes No

A consumer products testing group is evaluating two competing brands of tires, Brand 1 and Brand 2. Tread wear can vary considerably depending on the type of car, and the group is trying to eliminate this effect by installing the two brands on the same random sample of 10 cars. In particular, each car has one tire of each brand on its front wheels, with half of the cars chosen at random to have Brand 1 on the left front wheel, and the rest to have Brand 2 there. After all of the cars are driven over the standard test course for 20,000 miles, the amount of tread wear (in inches) is recorded, as shown in Table 1. Car Brand 1 Brand 2 Difference (Brand 1 - Brand 2) 1 0.362 0.235 0.127 2 0.343 0.368 -0.025 3 0.203 0.230 -0.027 4 0.335 0.247 0.088 5 0.207 0.221 -0.014 6 0.344 0.332 0.012 7 0.354 0.285 0.069 8 0.385 0.382 0.003 9 0.273 0.236 0.037 10 0.376 0.328 0.048 Table 1 Based on these data, can the consumer group conclude, at the 0.05 level of significance, that the mean tread wears of the brands differ? Answer this question by performing a hypothesis test regarding μd (which is μ with a letter "d" subscript), the population mean difference in tread wear for the two brands of tires. Assume that this population of differences (Brand 1 minus Brand 2) is normally distributed. Perform a two-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.) The null hypothesis: H0: _ The alternative hypothesis: H1: The type of test statistic: (Choose one) Z, t, Chi- square, and F The value of the test statistic: (Round to at least three decimal places.) The two critical values at the 0.05 level of significance: (Round to at least three decimal places.) _ and __ At the 0.05 level, can the consumer group conclude that the mean tread wears of the brands differ? Yes No

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Description: A consumer products testing group is evaluating two competing brands of tires, Brand 1 and Brand 2. Tread wear can vary considerably depending on the type of car, and the group is trying to eliminate this effect by installing the two brands on the sa

Calculus For The Life Sciences

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter1: Functions

Section1.2: The Least Square Line

Problem 8E

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Car	Brand 1	Brand 2	Difference (Brand 1 - Brand 2)
1	0.362	0.235	0.127
2	0.343	0.368	-0.025
3	0.203	0.230	-0.027
4	0.335	0.247	0.088
5	0.207	0.221	-0.014
6	0.344	0.332	0.012
7	0.354	0.285	0.069
8	0.385	0.382	0.003
9	0.273	0.236	0.037
10	0.376	0.328	0.048

Table 1

Based on these data, can the consumer group conclude, at the 0.05 level of significance, that the mean tread wears of the brands differ? Answer this question by performing a hypothesis test regarding μd (which is μ with a letter "d" subscript), the population mean difference in tread wear for the two brands of tires. Assume that this population of differences (Brand 1 minus Brand 2) is normally distributed.

Perform a two-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.)

The null hypothesis:	H₀: _____
The alternative hypothesis:	H_{1: ______}
The type of test statistic:	(Choose one) Z, t, Chi- square, and F

The value of the test statistic: (Round to at least three decimal places.)
The two critical values at the 0.05 level of significance: (Round to at least three decimal places.)	_____ and ____
At the 0.05 level, can the consumer group conclude that the mean tread wears of the brands differ?	Yes	No

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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Calculus For The Life Sciences

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9780321964038

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GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

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