Question

Test Statistic= 3,4, or 8

P-Value= 0.05,0.384, or 0.726

Conclusion= The distrabutions are different, The distrabutions are identical, or Reject H0

A car dealer was interested in comparing two brands of tires to see if they yielded different wear length (in thousands of miles). The dealer selected eight cars at random and used each of the brands of tires on each car. The wear length was recorded as follows: CAR BRANDA BRANDB 1 45 47 2 43 40 3 54 57 4 63 61 5 39 43 6 47 45 7 56 58 8 50 55 Use the sign test to see if the distribution of wear length is the same for both brands of tires. Use - 0.05. The null and alternate hypotheses are: Ho: The distributions of wear length are identical for the two brands of tires, and P[A exceeds B) =p -0.50. Ha: The distributions of wear length are not identical for the two brands of tires, and p 0.50. The test statistics is equal to [Select) The p-value for the test statistics is equal to Select Conclusion Select

Answer 1

For the given data, to test the above hypothesis using sign test,

Step 1: Computing the difference and noting the signs:

Brand A	Brand B	A-B	Sign
45	47	-2	-
43	40	3	+
54	57	-3	-
63	61	2	+
39	43	-4	-
47	45	2	+
56	58	-2	-
50	55	-5	-

Step 2: Add up the positives and negatives.

Here, the number of positives and negatives are 3 and 5 respectively. Here, sample size, n = 8.

Step 3: Test statistic here, is the number of positives or negatives, whichever is smaller. Here, we find the number of positives = 3.

Hence, test statistic = 3.

Step 4: Next, we may compute the p-value:

The given data follows binomial distribution with n = 8 and p = 0.5:

By using the binomial distribution formula:

P(x successes) x!(n-x)!

we can compute the probability of observing different numbers of successes during 8 trials:

$P(X=0)=\binom{8}{0}(0.5)^{0}(0.5)^{8}$

= 0.00390625

x	P(x successes)
0	0.00390625
1	0.03125
2	0.109375
3	0.21875
4	0.2734375
5	0.21875
6	0.109375
7	0.03125
8	0.00390625

Since, the p-value is the probability of observing a test statistic as or more extreme than that observed; we have observed 3 positives. Hence,

The two-sided p-value for the test is p-value = 2 x min [P(X$\leq$3),P(X$\geq$3)]

= 2 x [P(0) + P(1) + P(2) + P(3), (P(3) + P(4) + P(5) + P(6)+(P(7) + P(8)]

= 2 x [0.363, 0.855]

= 2 x 0.363

= 0.726

The p-value of the test = 0.726

Step 5: Since the p-value of the test = 0.726 > 0.05, we fail to reject the null hypothesis. The data does not provide sufficient evidence to support the claim that the two brands yielded different wear length.

We may conclude that the distribution of wire length is same for both brands of tires - The distributions are identical.