Consider the accompanying data on breaking load (kg/25 mm width) for various fabrics in both an unabraded (U) condition and an abraded (A) condition. Use the paired t test to test: H0: μD = 0 versus Ha: μD > 0 at significance level 0.01. (Use μD = μU-A.) Note: The data below is formatted such that you can copy and paste it into R. Fabric 1 2 3 4 5 6 7 8 U = c( 36.3, 55.0, 51.1, 38.8, 43.2, 48.8, 25.6, 49.5) A = c( 28.5, 20.0, 46.0, 34.5, 36.5, 52.5, 26.5, 46.5) Calculate the mean difference and standard deviation. d = sd = Compute the test statistic value. (Round your answer to three decimal places.) t = p-value = State the conclusion in the problem context. Fail to reject H0. The data does not suggest a significant mean difference in breaking load for the two fabric load conditions. Reject H0. The data does not suggest a significant mean difference in breaking load for the two fabric load conditions. Fail to reject H0. The data suggests a significant mean difference in breaking load for the two fabric load conditions. Reject H0. The data suggests a significant mean difference in breaking load for the two fabric load conditions.

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Answer:

see explaination

Step-by-step explanation:

Given a=0.01, the critical value is t(0.01, df=n-1=7) =3 (from student t table)

the critical region is one-sided

the rejection region is t>3

the test statistic value

t= mean difference / std.error

=7.35/4.1866

=1.756

Stating the conclusion in the problem context:

Fail to reject H0. The data does not suggest a significant mean difference in breaking load for the two fabric load conditions.

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