Question 1     

Suppose that you want to test the claim that the mean salary of veterinarians is $91,350. You collect salary information from a sample of 60 veterinarians and find that the mean salary in the sample is $90,350. The standard deviation is $5,000.


Part (a)            

At a level that would be understandable to a senior in high school, explain why it is necessary to carry out a hypothesis test for this claim (rather than rejecting the claim outright since the observed sample mean is clearly different from the claimed population mean).


Part (b)

Using a level of significance α of 0.1, test the null hypothesis H0: µ = $91,350 using the 6-step procedure.

Expert Answers

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We are asked to test the claim of the mean salary being $91,350 (`mu = 91350 ` ) using a sample of 60 salaries with a mean of $90,350 (` bar(x)=90350 ` ) and a population standard deviation of $5000 (s=5000.) ** I will recalculate if this is a sample standard deviation at the end.**

(a) The sample mean is not the sought after population mean, but we do not reject the claim outright. We would not expect the sample mean to be exactly 91350; there is some variability expected. If we were to do another sample of 60 salaries it is unlikely that the mean of this new sample would be 90350.

However, we do expect the sample means to be "close" to the population mean. In fact an application of the central limit theorem tells us that as the sample size grows, the means of the samples will be forced closer to the true population mean.

We use hypothesis testing to determine if the sample mean is outside of the expected range of sample means assuming that the population mean is actually 91350. Hypothesis tests take into consideration the size of the sample and how confident we want to be on accepting/rejecting the claim.

(b) You may need to adapt this to your 6-step process:

`H_0:mu=91350 ` This is the null hypothesis and our claim.

`H_1:mu != 91350 ` This is the alternative hypothesis.

With `alpha=.1 ` we compute the critical value(s):

Since the alternative is two-tailed (we allow for the actual mean to be above or below 91350) we look for the z-value with area .95 to the left (or .05 to the right.) Here z=1.645 (Some texts will use 1.64 or 1.65.) ((This number can be acquired from a standard normal table or from technology.))

The critical values are `z=+- 1.645 `

We now compute the test value. We have `n=60,sigma=5000,bar(x)=90350, mu=91350 ` so the test value is `z=(90350-91350)/(5000/sqrt(60))~~-1.549`

Since the test value is within the noncritical region (-1.549>-1.645) we do not reject the null hypothesis.

There is insufficient evidence to reject the claim that the mean salary is $91,350.


If this was the sample standard deviation we run a t-test: The sample is large enough that the critical values and test values are virtually the same with the same answer.

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