Question

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Unit 6    [Hypothesis Testing/Decision Making]       

For all Hypothesis\Decision Making Problems, please do the following:

            1.         Answer: Is the data discrete or continuous?

            2.        List the assumptions.

3.        State the null and alternate hypothesis.

            4          Write down the proper test statistic, Show all calculations, i.e. impute the numbers.                       .               Set up Decision rules. Show the critical value.    Reject or Fail to reject.  

            5.         State your conclusions.                        

6.5.1

Jacquemyn et al. conducted a survey among gynecologists-obstetricians in the Flanders region and obtained 300 responses. Of those responding, 90 indicated that they had performed at least one cesarean section on demand every year. Does this study provide sufficient evidence for us to conclude that less than 35 percent of the gynecologists-obstetricians in the Flanders region perform at least one cesarean section on demand each year? Let a = 05.

Answer 1

Solution:-

1) The data is discrete.

2)

The assumptions are:-

a) The value of np and n(1-p) are greater than 5.

b) The sampling distribution can be approximated as normally distributed.

c) The sampling method is simple random sampling.

d) Each sample point can result in just two possible outcomes. We call one of these outcomes a success and the other, a failure.

3)

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

Null hypothesis: P > 0.35
Alternative hypothesis: P < 0.35

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. 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 = 0.02754

z_critical = - 1.645

Critical region is z < - 1.645

-3 -2 zー-1.645 -1

z = (p - P) / S.D

z = - 1.816

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

4)

Interpret results. Since the z-value (- 1.816) lies in the rejection region, hence we have to reject the null hypothesis.

5) From the above test we have sufficient evidence in the favor of the claim that less than 35 percent of the gynecologists-obstetricians in the Flanders region perform at least one cesarean section on demand each year.