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I have a variable namely "Tourist satisfaction regarding ticket prices, transport cost a tourist have been charged by travel agent". I have created a 5 point scale.

1= Highly satisfied
2= little satisfied
3= Neutral
4= little dissatisfied
5= Highly dissatisfied

I am interested to test tourist satisfaction against mean satisfaction level (3=neutral). Can we apply T Test for that. Or is there any other test which is more appropriate.

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    $\begingroup$ Why do you think that 3 is the "mean satisfaction level"? You have to be more clear as to what exactly you want to test. $\endgroup$ – Zahava Kor Mar 30 '17 at 16:43
  • $\begingroup$ @ Zahava Kor, 3 is the neutral satisfaction which is indeed the mean of the satisfaction level. $\endgroup$ – Bilal Para Mar 31 '17 at 18:05
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Can you? Sure. But is it valid? That's harder to say. Note that Likert data cannot be normally distributed, which is an assumption of the t-test. However, if you have enough data, the sampling distribution of the mean might be 'normal enough'.

A different question (and a traditionally controversial one) is whether a mean would be a reasonable way to represent the information in Likert data (cf., What are good basic statistics to use for ordinal data?). This is a theoretical assumption that you are making and need to feel comfortable with and willing to defend. In your case, it isn't clear to me that you really care about the mean per se, so much as you want to be able to state that customers tend to be satisfied. Given that, I would use the Wilcoxon signed rank test to compare your sample to the value $3$ (which isn't really a hypothesized mean in this case). Another possibility that is even simpler is to count how many ratings are $>3$, and perform a binomial test against the null proportion $.5$. I think the latter will be especially easy to communicate to lay people (e.g., 'almost three quarters of our customers report being satisfied with our service'). Both of those options could be run as one-tailed tests.

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