Question

According to a study conducted two years ago by the Organisation for Economic Cooperation and Development (OECD), the average length of stay in a hospital was 16.2 days in Japan. A researcher in Japan suspects that the average length of stay in a hospital has actually increased in Japan since the study was released by the OECD two years ago. To test her suspicion, the researcher selected a random sample of 111 hospital patients in Japan and calculated the average length of hospital stay. The mean of the sample was found to be 16.26 days. Using a significance level of 2.5%, test the hypothesis that the mean length of stay in a hospital in Japan is more than 16.2 days. Assume that the standard deviation of the lengths of all hospital stays in Japan is known to be 2.11 days. Use the p-value method. State the null and alternative hypothesis for this test. H:27 H: V Determine if this test is left-tailed, right-tailed, or two-tailed. Oright-tailed Otwo-tailed O left-tailed Should the standard normal (2) distribution or Student's (t) distribution be used for this test? The Student's t distribution should be used O The standard normal (2) distribution should be used Determine the test statistic for the hypothesis test. Round the solution to two decimal places. Determine the p-value for the hypothesis test. Round the solution to four decimal places. Determine the appropriate conclusion for this hypothesis test. The sample data do not provide sufficient evidence to reject the alternative hypothesis that the average length of stay in a hospital in Japan is greater than 16.2 days and thus we conclude that the average length of stay in a hospital in Japan is likely greater than 16.2 days. The sample data provide sufficient evidence to reject the null hypothesis that the average length of stay in a hospital in Japan is 16.2 days and thus we conclude that the average length of stay in a hospital in Japan is likely longer than 16.2 days. The sample data provide sufficient evidence to reject the alternative hypothesis that the average length of stay in a hospital in Japan is greater than 16.2 days and thus we conclude that the average length of stay in a hospital in Japan is likely 16.2 days. The sample data do not provide sufficient evidence to reject the null hypothesis that the average length of stay in a hospital in Japan is 16.2 days and thus we conclude that the average length of stay in a hospital in Japan is likely 16.2 days.

Answer 1

Null hypothesis (Ho) : The average length of stay in hospital in Japan $(\mu)$ is 16.2 days.

$\mu = 16.2$

Alternative hypothesis (H1) : The average length of stay in hospital in Japan is more than 16.2 days

$\mu > 16.2$

Test is right tailed (as the alternative is greater than).

Since, the Population standard deviation is known, standard normal distribution (z) test should be used.

Test statistic is given by -

urlo 011

= Sample Mean = 16.26 days

T

Sample size = n = 111

Population standard deviation = $\sigma$ = 2.11 days

Specified value of the population mean under the null hypothesis = = 16.2 days

HO

Hence, the value of the test statistic will be -

$z = \frac{16.26-16.2}{2.11/\sqrt{111}}$

$= \frac{0.06}{0.2}$

= 0.30

P value = P(z > 0.30)

= Area under the z curve on the right side of z = 0.30

= 1 - P(z < 0.30)

The P(z < 0.30) can be obtained from the z table by finding the area corresponding to z = 0.3 at 0.00 probability and is equal to 0.6179

= 1 - 0.6179

= 0.3821

Level of significance = 2.5% = 0.025

As the P-value > level of significance, we may not reject the null hypothesis, hence, we conclude that the average length of stay in hospital in Japan is 16.2 days.

So, the conclusion will be -

The sample data do not provide sufficient evidence to reject the null hypothesis that the average length of stay in hospital in Japan is 16.2 days, and thus we conclude that the average length of stay in hospital in Japan is likely 16.2 days.

Hence, the fourth option is the correct option.