A survey of 100 similar-sized hospitals revealed a mean daily census in the pediatrics service of 27 with a standard deviation of 6.5. Researchers test whether these data provide sufficient evidence to indicate the mean is greater than 25. Use α=.05. Give 1. the hypotheses, 2. appropriate test, 3. decision rule, 4. calculated test statistic, and 5. conclusion with a comparison to the critical value or alpha.
Answers
Answer:
We conclude that the mean is greater than 25.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 25
Sample mean, = 27
Sample size, n = 100
Alpha, α = 0.05
Sample standard deviation, s = 6.5
First, we design the null and the alternate hypothesis
We use One-tailed(right) z test to perform this hypothesis.
Formula:
Putting all the values, we have
Now,
Since,
We reject the null hypothesis and accept the alternate hypothesis.
Thus, the mean is greater than 25.
Final answer:
The null hypothesis is the mean is equal to 25 and the alternative is that the mean is greater than 25. Using a one-sample t-test and 0.05 significance level, the calculated statistic results in rejection of the null hypothesis. Thus, there is sufficient evidence to suggest that the mean daily census in pediatrics is greater than 25.
Explanation:
1. The hypotheses for this scenario are that the null hypothesis (H0): the mean daily census in the pediatrics service is equal to 25, and the alternative hypothesis (H1): the mean daily census in the pediatrics service is greater than 25.
2. The appropriate test for this scenario would be a one-sample t-test, given that we have a sample mean, a population mean, a standard deviation, and we're examining a single group of hospitals.
3. The decision rule would be: if the p-value of our t-test is less than the significance level (α=.05), we reject H0 and accept H1.
4. The test statistic is calculated as follows: t = (Sample Mean-Population Mean)/(Sample Standard Deviation/ √number of observations), this would give us (27-25)/(6.5/√100) = 3.08.
5. Since 3.08 is greater than the critical value for a 0.05 significance level, we reject the null hypothesis. Therefore, there is sufficient evidence to conclude that the mean daily census in the pediatrics service is greater than 25.
Learn more about One-sample t-test here:
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Step-by-step explanation:
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Step-by-step explanation:
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