Hypothesis testing: significance i conducted a survey that aims to find out if Australians prefer Australian phones over Chinese phones.
My hypothesis states: Australians prefer Australian phones over Chinese phones
This was measured by the following question: 'If you had to choose between two identical phones, would you purchase the Australian or the Chinese phone?'
There are four choices:


*

*purchase Australian phone

*purchase chinese phone

*I'm indifferent

*i don't know


I guess it is not enough to only state how many people chose which option right?
I am unsure which statistical test to use to accept or reject the null hypothesis. I was thinking about a multiple linear regression, but then i didn't know how to code my variables (dependent: choices 1-4; independent: amount of people who chose the option?)
Please help me in figuring out which test is appropriate and how to code the variables given. I appreciate your help a lot!!
 A: Interviewing 200 randomly chosen Australians, suppose you got
87 favoring Australian phones, 77 favoring Chinese phones, and the
remaining 36 in one of the uninformative categories.
Then out of the $n = 164$ with relevant opinions you have $\hat p = 87/164 \approx 0.53 > 0.5$ in favor of Australian phones. Is that a large enough preference
for Australian phones to reject $H_0: p = .5$ against $H_a: p > .5,$
where $p$ is the population proportion of people with opinions who favor
Australian phones? 
Under $H_0,$ the number $X$ in favor of A has $X \sim \mathsf{Binom}(n=164,\, p=.5).$ The P-value is $P(X \ge 87) = 1 - P(X \le 86) = 0.2412 > .05,$
so you cannot reject $H_0$ at the 5% level of significance. [Computation in R.]
1 - pbinom(86, 164, .5)
[1] 0.2411531

As another example, suppose you interviewed $2000$ Australian subjects,
with 869 preferring A, 771 preferring C, and the rest in uninformative
categories. Again here you have $\hat p = 869/1640 \approx 0.53.$ However, now under $H_0,$ you have $X \sim \mathsf{Binom}(n=1640,\, p=.5)$
and $P(X \ge 869) = 0.0083 < 0.05,$ so you could reject $H_0$ at the 5% level of significance (even at the 1% level).
1 - pbinom(868, 1640, .5)
[1] 0.008291582

Notes: (1) The people without relevant opinions are removed from the sample before analysis. (2) It is not only the proportion in favor of A
that matters; the effective sample size (of people with opinions) must also be taken into account in finding the P-value.
(3) Plots of the relevant binomial PDFs are shown below. P-values are probabilities to the right of the vertical red lines.

(4) You should report numbers or percentages of respondents in
uninformative categories so that readers can have some idea how
important 'country of manufacture' is to Australians.
(5) Some statistical software programs include a 'binomial test' or 'test of a single proportion' as standard procedures. Some do exact computations and some use normal approximations. Here is a printout from Minitab (slightly edited for relevance) for my first example above, using an exact
computation.
Test and CI for One Proportion 

Test of p = 0.5 vs p > 0.5
                              Exact
Sample   X    N  Sample p   P-Value
1       87  164  0.530488     0.241

