Question

Q1. Perform a test of hypothesis to determine whether the average TiQ is lower than the industry standard of 2.5 minutes (150 seconds). Use a significance level of α=0.05.  

I just need to know the null and alternative hypothesis for this scenario and the decision rule. Since I can't upload the file, the decision rule can just be an example with dummy numbers.

Answer 1

The null hypothesis is a value which contains equality. The alternative hypothesis tells us the claim that is to be tested. It tells us whether the test is one tailed -left or right or if it is 2 sided.

Here we are testing for the average TiQ. So we want to test for population mean. We can either use t-test or z-test.

• If using z-test, we need the distribution to be normally distributed and we should have known population SD.
• If z-test is not possible then we can use student's t-test. For this we use sample SD. For this either the distribution should be normal or the sample size has to be large n> 30.

Test Stat = now this is for t-test. If z-test then replace Sx with pop SD.

is the null mean - the value which is tested.

For the critcal values

	One tailed	Two tailed
t-test
z-test

We use t-dist tables and normal tables for the values.

Criteria: Reject null hypothesis if |Test Stat| > |C.V. |

Another approach is p-value : probabiltiy of null hypothesis being true.

	One tailed	Two tailed
t-test	P( > |Test|)	2P( > |Test|)
z-test	P(Z >|test Stat|)	2P(Z > |test Stat|)

We use t-dist tables and normal tables for the values.

Criteria: Reject null hypothesis if p-value < level of significance.

In the question:

Perform a test of hypothesis to determine whether the average TiQ is lower than the industry standard of 2.5 minutes (150 seconds). Use a significance level of α=0.05.  

To test whether lower than null value

...inequality sign in null can be used or not is same

Since it is left the test stat is likely to be negative and so will the C.V, but using absolute gets you the same result.

If z-test =

= 1.6449

If t-test = ...........if n = 16 df = 16-1

= 1.7531

Reject null if |test Stat| > C.V.