If you wanted to prove to yourself that fewer than 7 people recognize the brand name, you could employ a 'one sample t-test'.
First, lets set up two hypotheses; $H_o$ and $H_a$.
We interpret the alternative hypothesis $H_a$ as what the 'researcher' believes, which in this case is you.
So, as the researcher you believe that the means are different. Not one greater than the other, just different, or 'unusual'.
Therefore,
$$H_a: \mu_A\ne \mu_B $$
Next, we can formulate $H_o$. Here $H_o$, which is the null hypothesis, is just the opposite of the alternative hypothesis $H_a$. So, the opposite of 'not equal' is really just, 'equal'.
Therefore,
$$H_o: \mu_A=\mu_B $$
Before we get into the nitty-gritty, lets think about what showing that a mean of $7$ is significantly different from $10.3$ would mean.
If we showed that $7$ was significantly different from $10.3$, that would mean what? - It would mean that everything less than $7$ is also significantly different.
Now we can use a couple of formulas to determine if the mean of the sample, which we will take as $7$, is significantly different from the accepted mean of $10.3$.
The two formulas of interest will be;
$$t = t_{\alpha , n-1}$$ and,
$$t.s.v. = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}$$
Here '$t$' represents our 'critical value', while '$t.s.v.$' represents our 'test statistic value'. (Hopefully you're familiar with what each of the variables above mean, and you're able to substitute the values appropriately.)
Once you evaluate $t$ and $t.s.v.$, you want to check which one is greater - which leads us to the final step in answering the question - the decision rule.
Here, our decision rule will be, "If $t.s.v.>t$, accept $H_o$." Or in other words, the accepted sample mean ($10.3$) is not different to the new sample mean ($7$).
The product rule is that: the probability for the intersection of independent events equals the product of the probabilities for each event.
However, liking football and liking sport are not independent events; since if you do not like sport you certainly will not like football. So this rule is inapplicable.
Indeed, liking football is a subset of liking sport. So the event of doing both is simply the event of liking football. That has a probability of $0.23$.
So by definition of conditional probability; $$\mathsf P(\text{Like(Football)}\mid\text{Like(A Sport)})=\dfrac{23}{84}$$
Best Answer
Let's assume that each adult answers independently. Let $X$ denote an indicator random variable, equal to 1, if an adult names the professional football his favorite sport, and 0 otherwise. We take the survey's result to mean that $\mathbb{P}(X=1) = p = 0.34$, and $\mathbb{P}(X=0) = 1-p = 0.66$. Thus $X$ is a Bernoulli random variable.
Asking the question of 130 adults, and counting the total of those favoring professional football, gives $$ N = X_1 + X_2 + \cdots+ X_{130} $$ The sum of independent Bernoulli random variables follows a binomial distribution, with parameters $n=130$ and $p=0.34$.
Thus $$ \mathbb{P}(N \leq 66) = \sum_{m=0}^{66} \binom{130}{m} (1-p)^{130-m} p^m = 0.99971 $$ and $$ \mathbb{P}(N > 31) = \sum_{m=32}^{130} \binom{130}{m} (1-p)^{130-m} p^m = 0.99168 $$
Using the central limit theorem approximation to the above probabilities gives a good agreement. Indeed, $\mu = 130 p = 44.2$ and $\sigma = \sqrt{130 p (1-p)} = 5.4$, so $$ \mathbb{P}(N \leq 66) \approx \Phi \left( \frac{66 - 44.2}{5.4} \right) = \Phi( 4.036) = 0.99973 $$ and
$$ \mathbb{P}(N > 31) \approx 1 - \Phi \left( \frac{31 - 44.2}{5.4} \right) = 1- \Phi( -2.444) = 0.992736 $$