# Use the sample data and a significance level of 0.01 to carry out a hypothesis test to test the administrator’s claim. Assume the number of beds filled per day is Normally distributed in the population. Use the hypothesis test structure as shown in lectures

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This image has been Flagged as inappropriate Click to unflag 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...

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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.

Approved by eNotes Editorial Team We are performing the one-sample t-test. It is a one-tailed test. The administrator is saying that the true occupancy level is less than 185

We find the mean and standard deviation of the 16 measurements and these are reported below:

x- bar = 175

s    = 14.28286

The null and alternative hypotheses are respectively:

Ho: mu = mu-naught     where mu-naught = 185

Ha : mu < mu- naught

• the standard error can be approximated by:

SE = s / sqrt( n )

•       SE = 14.28286/sqrt 16 = 3.57
• Degrees of freedom. The degrees of freedom (DF) is equal to the sample size (n) minus one. Thus, DF = n - 1.

•       DF  = 16 -1 = 15
• Test statistic. The test statistic is a t-score (t) defined by the following equation.

t = (x - μ) / SE

•              t =(175-185)/3.57 = - 2.80
• The critical t-score for d.f =15, one -tail and alpha = 0.01 is 2.602.

Conclusion: Since our test -statistic ( ignoring sign) of 2.80 is greater than the critical t-score of 2.602, it falls in the rejection region. We thus reject the null hypothesis that the occupancy level is 185 beds. The administrator is correct: at the 1% significance level the occupancy level is less than 185 beds.

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