Last statistics question of the semester for extra credit and I am done. Any help will be appreciated. Thank you

Question

The number of hours worked per week by medical residents is normally distributed with a standard deviation of 4.8 hours. We want to test the claim that the mean number of hours worked per week by medical residents is more than 60 hours. Based on a sample of 36 medical residents, how large the sample mean should be so that we fail to reject the claim at 5% significance level?

Answer 1

H₀: µ=60

H_a: µ>60

 

You want to fail to reject H₀ at the 5%-level, so the P-value needs to be at least 0.05.

Therefore, we want the sample means x such that z₀=(x-60)/(4.8/√36) results in a probability

P(z≥z₀) = 0.05

Use invNorm(0.95) and get z₀≤1.64.

Therefore,

(x-60)/(4.8/√36)≤1.64, or

 

x ≤ 61.3

 

Answer: If the sample mean is less than 61.3 hours, the significance level for this test is greater than 0.05 and you fail to reject H₀ at the 5%-level.

Answer 2

Hey Tito -- my apologies for the terminology in stats ... it reminds me of authorities using

Latin in the Dark Ages ... "std dev" of 4.8hr from the true center means that sample aves

will likely have 0.8hr s.d. from the true ave for sample sizes of 36 (4.8/sqrt 36 = 0.8).

 

I'm 95% sure that the true ave lies within +/- two s.d. of 0.8hr or within +/- 1.6hr ...

 

A sample mean of 61.6 hrs puts us two s.d. above the 60-hr threshold.

At 62 hrs, I'm almost certain the typical workweek exceeds 60 hrs ... Best wishes, sir :)