Quality Associates, Inc.
Case Study
Case Synopsis
In this case, we have Quality Associates, Inc. a consulting firm advising its client about sampling and statistical procedures that can be used to control their manufacturing process. Their client has offered samples to be analyzed, so they can quickly learn whether the process is operating satisfactorily or corrective actions needs to be taken.
The numbers given in the case were as follows: assumed population standard deviation is equal to .21, sample size is equal to 30 and the test value of the mean was 12. They also stated the two hypotheses to be tested: the null hypothesis that the population is equal to 12 and the alternative hypothesis that the mean is not equal to 12.
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At .0038, the p-value for sample 3 was less than the significance level set by the client. Sample 3 provides evidence to reject Ho and accept Ha. However, samples 1, 2 and 4 all have p-values larger than alpha, indicating that the process is working satisfactorily with a mean equal to 12. According to samples 1,2 and 4, corrective measures do not need to be taken.
A second approach to hypothesis testing is the critical value approach, which states to reject Ho if the z value is larger than z alpha/2 or smaller than –z alpha/2. Alpha was given at .01, alpha/2 is equal .005. Table 8.1, Values of Z alpha/2 for the most commonly used Z values, in the book states that for alpha/2 equal to .005, the z value is equal to 2.576. This is the z critical value for a two tailed test, outside of which lies the rejection area. Again, out of the 4 samples, only one had a critical value in the rejection area. The z value of sample 3, -2.8982, was smaller than the z critical value of -2.576. This leads to a rejection of Ho. Samples 1,2, and 4 all fall between the 2 critical values and provide evidence to not reject Ho.
Assumption
Based on the results of the hypothesis tests, both p-value approach and critical value approach, corrective action should be taken for sample 3. Samples 1,2 and 4 provide evidence that we cannot reject Ho, and therefore the
One-sample t-test are used in the parametric test which analyzes the means of populations. The t-test for independent groups are statistics that relates difference between treatment means to the amount of variability expected between any two samples of data within the same population (Hansen & Myers, 2012). Critical values are used in significant testing provide a range of t distribution that is used in whether a null hypothesis is rejected. Based on the data below as the level of significance is at .05, thus the critical values would fall under ±1.860 and the t value for this is 1.871 would suggest for the null to be rejected as it is greater than the critical value (Privitera, 2015, p. 267). Based on the population mean of 70 there was a mean difference of
Decision Rule: The calculated test statistic of -3.024 does fall in the rejection region of z<-1.645, therefore I would reject the null and say there is sufficient evidence to indicate mu<50.
When you perform a test of hypothesis, you must always use the 4-step approach: i. S1:the “Null” and “Alternate” hypotheses, ii. S2: calculate value of the test statistic, iii. S3: the level of significance and the critical value of the statistic, iv. S4: your decision rule and the conclusion reached in not rejecting or rejecting the null hypothesis. When asked to calculate p–value, S5, relate the p-value to the level of significance in reaching your conclusion.
Answer = I would accept the hypothesis. Because the experiement is replicable because only one test was done at a time and all steps were included.
(b) The value of Z with an area of 5% in the right tail, but not the sample mean.
“Hypothesis testing is a decision-making process for evaluating claims about a population” (Bluman, 2013, p. 398). This process is used to determine if you will accept or reject the hypothesis. The claim is that the bottles contain less than 16 ounces. The null hypothesis is the soda bottles contain 16 ounces. The alternative hypothesis is the bottles contain less than 16 ounces. The significance level will be 0.05. The test method to be used is a t-score. The test statistic is calculated to be -11.24666539 and the P-value is 1.0. The P-value is the probability of observing a sample statistic as extreme as the test statistic, assuming the null hypothesis is true. The T Crit value is 1.69912702. The calculations show there is enough evidence to support the claim that the soda bottles do
The critical value represents the point on the scale of test statistic value in which the null
We reject Ho if χ2 > χα2. At α=0.05, with 4 degrees of freedom, the critical value becomes χα2=9.488 (table E.4)
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4) Discuss the implications of changing the level of significance to a larger value. What mistakes or error could increase if the level of significance in
5) From calculations, computed z value is more than -1.65 and falls within Ho not rejected region. Ho is not rejected at α = 0.05 & α = 0.01 significance levels.
Conclusion : Fails to reject the null hypothesis. The sample does not provide enough evidence to support the claim that mean is significantly different from 12 .
For d1, t-statistic=- 2.5334, t-statistic > t-critical. Thus we reject Ho and d1 is significant.
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