Answer to Question #246043 in Statistics and Probability for boo

Question

An academic medical center surveyed all of its patients in 2002 to assess their satisfaction with
medical care. Satisfaction was measured on a scale of 0 to 100, with higher scores indicative of
more satisfaction. The mean satisfaction score in 2002 was 84.5. Several quality-improvement
initiatives were implemented in 2003 and the medical center is wondering whether the
initiatives increased patient satisfaction. A random sample of 125 patients seeking medical care
in 2003 was surveyed using the same satisfaction measure. Their mean satisfaction score was
89.2 with a standard deviation of 17.4. Is there evidence of a significant improvement in
satisfaction? Run the appropriate test at the 1% level of significance.

Answer 1

There is always some contention about the value of a parameter or the relationship between parameters. When parametric values are unknown we estimate them through sample values. If the Sample value is exactly the same as per our contention there is no high in accepting it. Any assumption about the parameter are probability function such a procedure is known as testing of hypothesis.

Given information

Sample size n = 125

sample mean $\bar{x} = 89.2$

Population mean μ =84.5

Sample standard deviation s = 17.4

Now we need to test z-test for single mean for 1% level of significance because it is a large sample test.

$H_0: μ = 84. 5 \\ H_1: \mu > 84.5$

Level of significance α = 0.01

Test statistic

$Z = \frac{\bar{x} - \mu}{s / \sqrt{n}} \\ Z = \frac{89.2-84.5}{17.4 / \sqrt{125}} = 3.02$

Read the Z-table value at 1% level of significance for Right tailed test is $Z_{tab} = 2.33$

Conclusion:

$Z > Z_{tab}$

We reject H₀.

Z falls in a rejection region.

There is significance difference between sample mean and population mean.