\"Durable press\" cotton fabrics are treated to improve their recovery from wrin
ID: 3327600 • Letter: #
Question
"Durable press" cotton fabrics are treated to improve their recovery from wrinkles after washing. Unfortunately, the treatment also reduces the strength of the fabric. The breaking strength of untreated fabric is normally distributed with mean 52 pounds and standard deviation 2.8 pounds. The same type of fabric after treatment has normally distributed breaking strength with mean 19 pounds and standard deviation 1.8 pounds. A clothing manufacturer tests 4 specimens of each fabric. All 8 strength measurements are independent. (Round your answers to four decimal places.)
Explanation / Answer
Solution:-
Untreated fabric is normally distributed with mean 52 pounds and standard deviation 2.8 pounds.
After treatment has normally distributed breaking strength with mean 19 pounds and standard deviation 1.8 pounds.
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 - 2 = 0
Alternative hypothesis: 1 - 2 0
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 1.664
DF = 6
t = [ (x1 - x2) - d ] / SE
t = 19.83
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 6 degrees of freedom is more extreme than -19.83; that is, less than -19.83 or greater than 19.83.
Thus, the P-value = less than 0.0001
Interpret results. Since the P-value (almost 0) is less than the significance level (0.10), we cannot accept the null hypothesis.
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