Question

In a trial designed to evaluate a standard test of the flammability of fabric used in children’s sleepwear, two different laboratories each tested ten pieces of the same type of fabric. The twenty test pieces were randomly assigned, ten to each laboratory. The response variable is the length of the char mark made when the fabric piece was held over a flame for a specified time period. The following data were obtained:
Laboratory Char mark lengths (cm)
A 2.9 3.1 3.1 3.7 3.1 4.2 3.7 3.9 3.1 3.0
B 2.7 3.4 3.6 3.2 4.0 4.1 3.8 3.8 4.3 3.4
The sample means and standard deviations are as follows:

Laboratory	Mean	Standard deviation
A	3.38	0.452
B	3.63	0.474

(a) State null and alternative hypotheses for comparing the mean char mark lengths for the two laboratories, with a view to determining whether there is a systematic difference between the two.
(b) Using an appropriate test, compute the test statistic, give the P-value and state your conclusion.
(c) If the trial were to be repeated with the aim of reducing the width of a 95% confidence interval for the mean difference by a factor of two, what sized samples should be tested?

Answer 1

The hypothesis are stated as below: Null H_o = > mu_A = mu_B H_a = > mu_A notequalto mu_B Here, mu denotes the mean char mark length calculating test statistic: t = m_B-m_A/squareroot s^2_B/10 + s^2_A/10 = 3.63 - 3.38/squareroot (0.474)^2/10 + (0.452)^10/10 = 1.207 Fegrees of freedom, df = 10 + 10 - 2 = 18 The corresponding p-value is p = 0.243 (because this is a two tailed test) since p > 0.05 so we cannot reject the null hypothesis width of a 95% C.I. = 2 times margin of error = 2 times (2.1 times standard error) = 4.1 times squareroot s^1_A + s^2_B/10 so, width of C.I alpha 1/squareroot samplize so, if yu have sample size 4 times longer, tthen width gets reduce by half. So the sample size must be increased by a factor of 4, so that the width of interval decreases by a factor of two. So sample size to be tested this time should be: 20*4 = 80

So the sample size must be increased by a factor of 4, so that the width of interval decreases by a factor of two.

So sample size to be tested this time should be: 20*4 = 80