Dr. Elliott suspects a difference in the quality of a material her lab purchased
ID: 3236899 • Letter: D
Question
Dr. Elliott suspects a difference in the quality of a material her lab purchased from two different suppliers. The following data were obtained on the amount of impurity (in mg) per package from the suppliers. For supplier A, 65 samples were taken with a mean of 1.42 and a standard deviation of 0.19; for supplier B, 62 samples were taken with a mean of 1.47 and a standard deviation of 0.12. Dr. Elliott wants to test if the difference between the two means is statistically significant at the 99% confidence level (alpha = 0.01), assuming the population variances are equal. a) Write the null hypothesis and the alternative hypothesis in equation form. b) Compute the test statistic. Find the critical value. Should Dr. Elliott reject or not reject the null hypothesis? Explain why based on comparing the value of the test statistic to the critical value. c) Compute the p value. State whether or not it is significant at the 99% confidence level and why. d) Can Dr. Elliott conclude with 99% confidence that the quality of the material purchased from the two suppliers may be the same?Explanation / Answer
Solution:-
The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 - 2 = 0, i.e., the material purchased from the two suppliers may be the same.
Alternative hypothesis: 1 - 2 0, i.e., the material purchased from the two suppliers may not be the same.
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 0.01(as we are given 99% confidence interval). 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 = sqrt[(0.192/65) + (0.122/62] = 0.0281
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (0.192/65 + 0.122/62)2 / { [ (0.192 / 65)2 / (64) ] + [ (0.122 / 62)2 / (61) ] }
DF = 108.764 or 109
t = [ (x1 - x2) - d ] / SE = [ (1.42 - 1.47) - 0 ] / 0.0281 = -7/3.51 = -1.779
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 109 degrees of freedom is more extreme than -1.779; that is, less than -1.779 or greater than 1.779.
We use the t Distribution Calculator to find P(t < -1.779)
The P-Value is 0.078027.
The result is not significant at p < 0.01.
Interpret results. Since the P-value (0.078) is greater than the significance level (0.01), we can accept the null hypothesis.
Conclusion. Do not reject null hypothesis. i.e., we conclude with 99% confidence that the quality of the material purchased from the two suppliers may be the same.
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