Question

In: Statistics and Probability

A Patent Medicine Company supervisor assumes that the bottling machine is operating properly if only 5...

A Patent Medicine Company supervisor assumes that the bottling machine is operating properly if only 5 percent of the processed bottles are not full. A random sample of 100 bottles had 7 bottles that weren’t full. Using a significance level of .01, conduct a test to see if the machine is operating properly. 10. According to Management Accounting, salary figures for certified management accountants (CMAs) who are in the field less than 1 year are normally distributed with a mean of $31,129. A random sample of 15 firstyear CMAs in Denver produces a mean salary of $32,279, with a standard deviation of $1,797. Test the hypothesis that the mean for all Denver firstyear CMAs is not equal to $31,129. Use the .05 level of significance.

1. State the null hypothesis H0: _____________

2. State the alternative hypothesis H1: _____________

3. What is the test statistic used for the test (z or t)

4. State the significance or alpha (?) level

5. Determine the p-value.

6. Do you or do you not reject the null hypothesis? Why?

7. Write a clear conclusion using a complete sentence.

Solutions

Expert Solution

Solution:-

9)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P = 0.05
Alternative hypothesis: P 0.05

Note that these hypotheses constitute a two-tailed test.  

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation (S.D) and compute the z-score test statistic (z).

S.D = sqrt[ P * ( 1 - P ) / n ]

S.D = 0.02179
z = (p - P) /S.D

z = 0.92

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 0.92 or greater than 0.92.

Thus, the P-value = 0.358

Interpret results. Since the P-value (0.358) is greater than the significance level (0.01), we cannot reject the null hypothesis.

Do not reject H0.

From the above test we have sufficient evidence in the favor of the claim that the machine is operating properly.


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