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.
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
2 answers
Am need solves