Question

4. The length of stay (in days) from 100 randomly selected hospital patients are presented in the table below. Suppose we want to test the hypothesis that the population mean length of stay at the hospital is less than 5 days. Conduct hypothesis using =.05.       

2	3	8	6	4	4	6	4	2	5
8	10	4	4	4	2	1	3	2	10
1	3	2	3	4	3	5	2	4	1
2	9	1	7	17	9	9	9	4	4
1	1	1	3	1	6	3	3	2	5
1	3	3	14	2	3	9	6	6	3
5	1	4	6	11	22	1	9	6	5
2	2	5	4	3	6	1	5	1	6
17	1	2	4	5	4	4	3	2	3
3	5	2	3	3	2	10	2	4	2

Answer 1

Solution:-

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

Null hypothesis: u = 5
Alternative hypothesis: u < 5

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.367756
DF = n - 1

D.F = 99
t = (x - u) / SE

t = - 1.28

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of - 1.28.

Thus the P-value in this analysis is 0.102.

Interpret results. Since the P-value (0.102) is greater than the significance level (0.01), we cannot reject the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that the population mean length of stay at the hospital is less than 5 day.