Hypothesis Testing for Mean (small samples)

[exbklyngirl]

Metro Bank claims that the mean wait time for a teller during peak hours is less than 4 minutes. A random sample of 20 wait times has a mean of 2.6 minutes with a sample standard deviation of 2.1 minutes. Assume the distribution is normally distributed.
A. Use the critical value t method from the normal distribution to test for the population mean. Test the company's claim at the level of significance = 0.05.

The thing I figured is the following by I really don't understand how or what formula to use to get the rest of the answers. Please help.

A. Ho: µ=0.04
Ha:µ?0.04
A1. =0.05 or from the z score it would equal -1.64 ? not too sure on that.
A2. test statistics: two-tailed test ??
A3. P-value or critical Zo or t
A4. Rejection Region:
A5. Decision:
A6. Interpretation:

Would I be using the formula? z=x=µ/??n

 

[exbklyngirl]

Sorry about the formula but it should be:

z=x-µ/Ó?n

hope that is the way it should read.

 

[galactus]

\(\displaystyle H_{0}: \;\ {\mu}\geq 4\)

\(\displaystyle H_{a}: \'\ {\mu}<4 \;\ \text{claim}\)

Using \(\displaystyle t=\frac{(x-{\mu})\sqrt{n}}{\sigma}\), we get a test stat of:

\(\displaystyle t=\frac{(2.6-4)\sqrt{20}}{2.1}=-2.98\)

Look up the critical value in a t table under 19 d.f. and we see t=-1.729

Since -2.98 < -1.729, it is in the rejection region and we reject the null hypothesis.

\(\displaystyle \usepackage{cancel}\cancel{H_{0}: \;\ {\mu}\geq 4}\)

\(\displaystyle H_{a}: \;\ {\mu}<4 \;\ \text{claim}\)

The only thing left is the alternate hypothesis. This means the banks claim is valid at the .05 level.

 

[Deleted member 4993]

Just remember, for sample size less than 30 (n<30) , generally t-test is used to compare "average".