Answer to Question #210176 in Statistics and Probability for Ateca

Question

Employees of a large corporation are concerned about the declining quality of medical services provided by their group health insurance. A random sample of 100 office visits by employees of this corporation to primary care physicians during the year 2017 found that the doctors spent an average of 19 minutes with each patient. This year a random sample of 108 such visits showed that doctors spent an average of 15.5 minutes with each patient. Assume that the standard deviations for the two populations are 2.7 and 2.1 minutes, respectively. Using the 2.5% level of significance, can you conclude that the mean time spent by doctors with each patient is lower for this year than for 2017? To draw your conclusion, state the hypotheses and identify the claim, find the critical value(s), label the acceptance and rejection region, calculate the test value and summarize the results.

Answer 1

Let 1 denote parameters and statistic for 2017 and 2 denote the same for this year. Since population standard deviations are known Z-score is used.

$H0:\mu_1=\mu_2$

$Ha:\mu_1>\mu_2$ (Claim)

Critical value $=z_{0.025}=1.96$ (right tailed)

$\bar{x_1}=19$

$\sigma_1=2.7$

$n_1=100$

$\bar{x_2}=15.5$

$\sigma_2=2.1$

$n_2=108$

Test statistic.

$\displaystyle{z}=\frac{(\overline{x}_1-\overline{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{(\sigma_1)^2}{n_1}+\frac{(\sigma_2)^2}{n_2}}}$

$\displaystyle{z}=\frac{19-15.5}{\sqrt{\frac{(2.7)^2}{100}+\frac{(2.1)^2}{108}}}=10.378$

Since 10.378 fall in the rejection region, (test statistic is greater than the critical value) we reject the null hypothesis.

There is enough evidence to support the claim that the mean time spent by doctors with each patient is lower for this year than for 2017