It would be easier to answer directly if you had provided data. I will answer
by illustrating with fictitious data. If issues arise with your data that are
not covered here, please edit your question to provide pertinent information, and maybe someone will respond.
Suppose the sample size is $n = 200$ subjects, and that the sample
distribution of the differences
d between Likert scores for Apples and Pears
is as follows.
cor(appl, pear, meth="spearman")
On average, there are higher scores for Apples, and people who like one
seem to like the other (Spearman rank correlation is $0.78.)$
d = appl - pear
-2 -1 0 1 2
15 29 70 59 27
A paired Wilcoxon signed rank test is a test on the differences 'd'. There are
many tied values, but with $n$ as large as 200, the P-value is an approximation
that deals appropriately with ties. The null hypothesis that the population is centered at $0$ is strongly rejected with P-value about $0.001 < 0.05 = 5\%.$
Wilcoxon signed rank test
with continuity correction
V = 5582, p-value = 0.001299
alternative hypothesis: true location is not equal to 0
Even with $n=200$ it seems quite a reach to view a discrete distribution with
only five integer values as normal, but for the record a paired t test also
gives a very small P-value.
Finally, of the $200 - 70 =130$ subjects who expressed a clear preference, only $14+29=44$ preferred pears. So a 2-sided exact binomial test rejects equal preferences for apples and pears with P-value $0.0003 < 0.05 = 5\%.$
pbinom(44, 130, .5)*2
Note: Here is how I used R to sample the fictitious data used in the samples above.
appl = sample(1:7, 200, rep=T, p=c(1,2,3,4,5,5,4) )
dif = sample(-2:2, 200, rep=T, p=c(1,1,2,2,1))
pear = appl - dif
pear[pear>7]=7; pear[pear < 1]=1