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Case study, Pages 4 (957 words)

In this case, we have Quality Associates, Inc. a consulting company advising its client about tasting and analytical procedures that can be used to manage their production process. Their client has used samples to be evaluated, so they can rapidly learn whether the process is running adequately or corrective actions requires to be taken.

The numbers given in the case were as follows: assumed population standard variance amounts to. 21, sample size is equivalent to 30 and the test worth of the mean was 12.

They likewise stated the 2 hypotheses to be tested: the null hypothesis that the population is equivalent to 12 and the alternative hypothesis that the mean is not equivalent to 12. This indicates a 2 tailed tests to figure out whether to turn down the null hypothesis. The 4 supplied sample sizes each included 30 observations, showing a typical distribution and z test statistics. Method The very first concern required performing a hypothesis test for each sample at the. 01 level of significance.

Based upon the test, identify if any corrective actions require to be taken. There are two methods to hypothesis screening, the p-value approach and the crucial value approach. The first action for the p-value method was to determine the mean for each sample. In order, they were: 11. 9587|12. 0287|11. 8890|12. 0813|Next, was to compute basic mistake, by utilizing the formula sigma divided by the square root of n. This came out to be. 0383. To discover the z test fact subtract the test worth of 12 from the sample mean and divide by the basic error.

The z test statistic for each sample were as follows: -1.

0966| 0. 7493| -2. 8982| 2. 1227| The 1 tail p-value could then be found by using the normsdist function in Excel. This function indicates probability to the left of the value, so positive numbers were subtracted from 1. Since this is a two tailed test, the values were multiplied by two to find the actual p-value. 0. 2728| 0. 4536| 0. 0038| 0. 0338| The rule for the p-value two tail test is to reject Ho if the p-value is smaller than or equal to alpha, the alpha in this case being . 01. At . 038, 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 . 1, 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 client can assume that the process is operating satisfactorily. The second question asked to compute the standard deviation for each individual sample. Using the stdev function in Excel, the standard deviation for each sample are as follows: 0. 2204| 0. 2204| 0. 2072| 0. 2061| Assumption

Based on the standard deviations calculated for each sample, the assumption of . 21 for the population standard deviation appears reasonable. An average of the 4 individual sample standard deviations is equal to . 2135, which can be rounded down to . 21. The third question asked to compute limits for the sample mean equal to 12. Condition was that as long as a new sample mean is within those limits, the process will be considered to be operating satisfactorily. If X exceeds the upper limit or if the X is below the lower limit, corrective actions will be taken.

The formula for calculating upper and lower control limits is Using x bar equal to 12, z alpha/2 equal to 2. 576 and the standard error equal to . 0383, the upper limit was equal to 12. 0987 and the lower limit was equal to 11. 9013. Assumption Based on the upper (12. 0987) and lower (11. 9013) control limits calculated for a mean equal to 12, sample 3 falls outside the control limit with a mean of 11. 8890. Because the mean exceeds the lower limit, it indicates that corrective action needs to be taken.

Assumptions Increasing the level of significance to a larger value will lead to rejecting the null hypothesis more often. If the level of significance is increased to . 05, both samples 3 and 4 will provide evidence to reject the null hypothesis instead of just sample 3. This means that the client is more willing to make a Type I error, mistakenly rejecting Ho when it is true. This means that they run the risk of having to stop their manufacturing process to take corrective action more often.