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.                        

6.3.1    

Suppose it has been established that for a certain type of client (fair health-Group A) the average length of a home visit by a public health nurse is longer than for a second type of client (poorer health-Group B). If nurses on staff randomly visit 36 clients from Group A and record the average time as 30 minutes with a standard deviation of 15 minutes, and the average home visit for Group B (49 visits) is 45 minutes long with a standard deviation of 20 minutes, test by hypothesis that the two groups of patients have different home visit times. Assume data is normally distributed.

Answer 1

Solution:-

1) The data is continuous.

2)

The assumptions are:-

a) The sample sizes are sufficiently large, n > 30.

b) The population standard deviation is unknown.

c) The population distribution is normally distributed.

3)

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

Null hypothesis: u₁ = u ₂
Alternative hypothesis: u₁ u ₂

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. 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 = 3.7965
DF = 83

t_critical = + 1.989

Critical region is -1.989 > t > 1.989

t = [ (x₁ - x₂) - d ] / SE

t = - 3.95

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.

4)

Interpret results. Since the t-value (- 3.95) 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 two groups of patients have different home visit times.