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I'm trying to figure this out and either I am completely missing the point or I am in fact correct. (I am a student).

The null hypothesis is "There is no significant difference between time taken to complete a booking between the two group, new users and previous users".

It is my understanding that to reject this I need a 95% CI that doesn't lie within 0. However from the image below it does {-25.47, 4.65}. But the p value is .172 which accepts the null. I cant seem to find online what I am not understanding? Do i accept it from the 95% CI or have I got something misunderstood?

Thanks in advance!

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  • $\begingroup$ I think you need to include the self-study tag on this one mate. $\endgroup$
    – André.B
    Commented May 9, 2019 at 20:48
  • $\begingroup$ Properly, your null hypothesis should be a statement about a population (as should your alternative); the word "significant" doesn't belong in a hypothesis. $\endgroup$
    – Glen_b
    Commented May 10, 2019 at 1:33

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There is no significant difference between the two groups, which can be inferred from the fact that 0 is included in the confidence interval. Therefore, we have no evidence against the null hypothesis, which is that there is no difference between the groups.

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  • $\begingroup$ Thanks - I've also just realized I was thinking that for 5% it was a p less that 0.05 and not 0.5! You may be able to tell math is not my strong point! $\endgroup$
    – George
    Commented May 9, 2019 at 20:54
  • $\begingroup$ No worries! Glad I could help - we all start somewhere. $\endgroup$
    – André.B
    Commented May 9, 2019 at 20:59
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I agree with what André stated. There is not a significant difference between groups. Both the confidence interval (which contains zero) and the result of the t test, t(56) = -1.286, p = .204 indicate a failure to reject the null hypothesis.

I do want to add one thing, with a t-test, we either reject or fail to reject the null hypothesis. We are not able to accept it. It may sound a bit particular, but it is an important distinction to make.

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