A medical board claim that the mean number of hours worked per week by surgical faculty who teach at an academic institution is more than 60 hours. A random sample of 7 surgical facility has a mean hours worked per week of 70 hours and a standard deviation of 12.5 hours. At a=0.05, do you have enough evidence to support boards claim?
State Null, alternative hypothesis, identify claim
Find Critical values, critical region
Try a t-test since your sample size is rather small.
Ho: µ = 60 -->null hypothesis
Ha: µ > 60 -->alternative hypothesis
Formula:
t = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
sample mean = 70
population mean = 60
standard deviation = 12.5
sample size = 7
Plug in the values and calculate the t-test statistic.
Find the critical value for a one-tailed test at .05 level of significance using 6 for degrees of freedom (df = n - 1). Use a t-table. Compare to your t-test statistic calculated above. If the t-test statistic exceeds the critical value from the table, reject the null and conclude µ > 60. If the t-test statistic does not exceed the critical value from the table, do not reject the null.
I hope this will help.
Am need solves
To determine whether there is enough evidence to support the medical board's claim, we need to conduct a hypothesis test. Here are the steps involved:
Step 1: State the Null and Alternative Hypotheses.
The null hypothesis (H0) assumes that the mean number of hours worked per week by surgical faculty is less than or equal to 60 hours, i.e., μ ≤ 60.
The alternative hypothesis (Ha) assumes that the mean number of hours worked per week is greater than 60 hours, i.e., μ > 60.
Step 2: Identify the Claim.
The claim made by the medical board is that the mean number of hours worked per week by surgical faculty is more than 60 hours.
Step 3: Determine the Significance Level.
The significance level (denoted as α) is given as 0.05 in the question.
Step 4: Find the Critical Values or Critical Region.
Since the sample size is small (n < 30) and the population standard deviation (σ) is unknown, we will use a t-test. The critical values or critical region will depend on the degrees of freedom and the significance level chosen.
Degrees of freedom (df) = n - 1 = 7 - 1 = 6
For a one-tailed t-test, at α = 0.05 and df = 6, the critical value can be found using a t-table or a statistical software. In this case, the critical value is approximately 1.943.
Critical Region: The critical region will be the right tail of the t-distribution, with t > 1.943.
Step 5: Calculate the Test Statistic.
The test statistic, t, can be calculated using the formula:
t = (sample mean - hypothesized population mean) / (sample standard deviation / √n)
Plugging in the given values, t = (70 - 60) / (12.5 / √7) ≈ 2.11
Step 6: Make a Decision.
If the test statistic falls within the critical region (t > 1.943), we reject the null hypothesis and have enough evidence to support the claim. Otherwise, if the test statistic falls outside the critical region, we fail to reject the null hypothesis.
The calculated test statistic, t = 2.11, falls within the critical region t > 1.943. Therefore, we reject the null hypothesis and conclude that there is enough evidence to support the claim made by the medical board.