A consumer products testing group is evaluating two competing brands of tires, Brand 1 and Brand 2. Though the two brands have been comparable in the past, some technological advances were recently made in the Brand 2 manufacturing process, and the consumer group is testing to see if Brand 2 will outperform Brand 1. 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 12 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 0.366 0.319 0.047 2 0.364 0.307 0.057 3 0.316 0.359 -0.043 4 0.277 0.298 -0.021 5 0.355 0.326 0.029 6 0.268 0.306 -0.038 7 0.355 0.375 -0.020 8 0.362 0.361 0.001 9 0.391 0.266 0.125 10 0.381 0.397 -0.016 11 0.394 0.337 0.057 12 0.323 0.356 -0.033 Table 1 Based on these data, can the consumer group conclude, at the 0.10 level of significance, that the mean tread wear of Brand 1 exceeds that of Brand 2? 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 one-tailed test. The null hypothesis: H0: The alternative hypothesis: H1: The type of test statistic: (Choose one)ZtChi squareF The value of the test statistic:(Round to at least three decimal places.) The critical value at the 0.10 level of significance:(Round to at least three decimal places.) At the 0.10 level, can the consumer group conclude that the mean tread wear of Brand 1 exceeds that of Brand 2? Yes No

A consumer products testing group is evaluating two competing brands of tires, Brand 1 and Brand 2. Though the two brands have been comparable in the past, some technological advances were recently made in the Brand 2 manufacturing process, and the consumer group is testing to see if Brand 2 will outperform Brand 1. 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 12 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 0.366 0.319 0.047 2 0.364 0.307 0.057 3 0.316 0.359 -0.043 4 0.277 0.298 -0.021 5 0.355 0.326 0.029 6 0.268 0.306 -0.038 7 0.355 0.375 -0.020 8 0.362 0.361 0.001 9 0.391 0.266 0.125 10 0.381 0.397 -0.016 11 0.394 0.337 0.057 12 0.323 0.356 -0.033 Table 1 Based on these data, can the consumer group conclude, at the 0.10 level of significance, that the mean tread wear of Brand 1 exceeds that of Brand 2? 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 one-tailed test. The null hypothesis: H0: The alternative hypothesis: H1: The type of test statistic: (Choose one)ZtChi squareF The value of the test statistic:(Round to at least three decimal places.) The critical value at the 0.10 level of significance:(Round to at least three decimal places.) At the 0.10 level, can the consumer group conclude that the mean tread wear of Brand 1 exceeds that of Brand 2? Yes No

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

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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	0.366	0.319	0.047
2	0.364	0.307	0.057
3	0.316	0.359	-0.043
4	0.277	0.298	-0.021
5	0.355	0.326	0.029
6	0.268	0.306	-0.038
7	0.355	0.375	-0.020
8	0.362	0.361	0.001
9	0.391	0.266	0.125
10	0.381	0.397	-0.016
11	0.394	0.337	0.057
12	0.323	0.356	-0.033

Table 1

Based on these data, can the consumer group conclude, at the

0.10

level of significance, that the mean tread wear of Brand 1 exceeds that of Brand 2? Answer this question by performing a hypothesis test regarding

μd

(which is

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 one-tailed test.


The null hypothesis:	H0:
The alternative hypothesis:	H1:
The type of test statistic:	(Choose one)ZtChi squareF

The value of the test statistic: (Round to at least three decimal places.)
The critical value at the 0.10 level of significance: (Round to at least three decimal places.)
At the 0.10 level, can the consumer group conclude that the mean tread wear of Brand 1 exceeds that of Brand 2?	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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