A hospital claims that the mean waiting time at its emergency room is 25 minutes. A random sample of 16 patients produced a mean wait time of 27.5 minutes with a standard deviation of 4.8 minutes. Use the 1% level of significance to test is the wait time is different from 25 minutes. Assume that wait time is normally distributed. Use the 5 step procedure.
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To test whether the wait time at the hospital's emergency room is different from 25 minutes, we can use the 5-step procedure for hypothesis testing at a 1% level of significance. Here are the steps:
Step 1: State the null and alternative hypotheses.
The null hypothesis (H₀) is that the mean wait time is equal to 25 minutes.
The alternative hypothesis (H₁) is that the mean wait time is different from 25 minutes.
H₀: µ = 25
H₁: µ ≠ 25
Step 2: Set the significance level.
The significance level (α) is given as 1%. This means we are willing to make a Type I error (rejecting the null hypothesis when it is true) in only 1% of the cases.
α = 0.01
Step 3: Calculate the test statistic.
The test statistic for comparing means when the population standard deviation (σ) is unknown is the t-score. We can calculate it using the formula:
t = (sample mean - hypothesized mean) / (sample standard deviation / √sample size)
Given:
Sample size (n) = 16
Sample mean (x̄) = 27.5
Sample standard deviation (s) = 4.8
Hypothesized mean (µ₀) = 25
t = (27.5 - 25) / (4.8 / √16)
t ≈ 2.08 (rounded to two decimal places)
Step 4: Determine the critical value(s).
To determine the critical value(s), we need to consult the t-distribution table or use statistical software. Since we are conducting a two-tailed test (µ ≠ 25), we need to find the critical values that correspond to a significance level of 0.005 on each tail.
Using a t-table or software, we find that the critical value for a two-tailed test with α = 0.01 and degrees of freedom (df) = n - 1 = 16 - 1 = 15 is approximately ±2.947.
-2.947 < t < 2.947
Step 5: Make a decision and interpret the result.
If the calculated t-value falls within the range of the critical values, we fail to reject the null hypothesis. If the calculated t-value falls outside the range, we reject the null hypothesis in favor of the alternative hypothesis.
Since 2.08 does not exceed the critical values of ±2.947, we fail to reject the null hypothesis. There is not enough evidence to conclude that the mean wait time at the hospital's emergency room is different from 25 minutes.
Therefore, based on the given data and using a 1% level of significance, we do not have sufficient evidence to support the claim that the wait time is different from 25 minutes.