I'm so confused with Null hypothesis and Alternative hypothesis in this solution So the question says "A company that claims the average time a customer waits on hold is less than $5$ minutes. A sample of $35$ customers have an average wait time of $4.78$ minutes . Assume the population standard deviation for the wait time is $1.6$ minutes. Test the company's claim"
So to me the null hypothesis is the status quo therefore $H_0\lt 5$ whereas the alternative hypothesis $H_1$ should be $H_1\ge5$
However the official answer to this solution is completely opposite it's stated $H_0\ge5$ , $H_1\lt5$ 
What am I missing here?
 A: You aren't missing anything.
Your answer is mathematically completely correct.
Unfortunately, there are people out there, who try to set up fixed rules when setting up the hypotheses like "status quo is always $H_0$ and claim is always $H_1$" or the other way round. Fact is, there is no general fixed rule of this type.
In your case, you can carry out a hypothesis test both for 


*

*$H_0: \mu \lt 5$ vs. $H_1: \mu \geq 5$ and

*$H_0: \mu \geq 5$ vs. $H_1: \mu \lt 5$


Important is, that the hypotheses are designed in a way, such that a meaningful test statistic can be calculated.
For example, when testing data for following a certain distribution using $\chi^2$-tests, $H_0$ is exactly the claim.
Nevertheless, in many cases, there is a status quo which is often set to be $H_0$. The claim is then $H_1$ (something has changed, the status quo is probably not valid anymore). 
Then, you test whether there is enough statistical evidence which supports that $H_0$ is rather improbable. $H_0$ is considered to be improbable if the test value of the sample is improbable under the assumption that $H_0$ is true given a certain significance level $\alpha$.
A: In statistics, the alternative hypothesis is what we wish to show. In this case, the company wishes to show that the average time a customer waits on hold is less than $5$ minutes.
We have
$$\frac{4.78-5}{1.6/\sqrt{35}}\sim \mathsf N(0,1)$$
Since $$\Phi\left(\frac{4.78-5}{1.6/\sqrt{35}}\right)\approx\Phi(-0.813)\approx0.208$$
we fail to reject the null hypothesis at $\alpha=0.05$. We do not have significant evidence that the average time a customer waits on hold is less than $5$ minutes.
