For this numerical, we can use the t-test to determine if the null or alternate hypothesis is true. The null hypothesis is the administrator's claim that at least 185 beds are filled on any given day. The alternate hypothesis is the suspicion of board of directors that the number of filled beds is less than 185. Thus, this is a left-tailed test.
that is, `H_0 : m = 185 and H_1 : m < 185`
One can calculate the required parameters using a calculator or any spreadsheet program.
The mean of the given data is 175 and the standard deviation is 14.283. Since the given significance level is 0.01, the confidence interval is 98% and the corresponding z-score is - 2.33.
If the z-level is less than the z-score, we will reject null hypothesis and accept alternate hypothesis. If the z-level is more than z-score than the test is not statistically significant.
z-level =`(175-185)/(14.283/sqrt(16)) = -2.8`
Since the z-level is less than the z-score of -2.33, we reject the null hypothesis (administrator's claim) and accept the alternate hypothesis that the occupancy was less than 185 beds per day.
Hope this helps.
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