Solve the problem that involves Hypothesis testing and Bin (n,p)

[chwala]

Homework Statement

See attached

Relevant Equations

Hypothesis tests and Binomial distribution

$p(x≤11)=1-[p(x=12)+p(x=13)+p(x=14)+p(x=15)]$
$p(x≤11)=1-[0.128505+0.266895+0.3431518+0.205891]=1-0.9444428=0.0556$
$⇒p(x>11)=0.9444$
$p(x≤10)=1-[0.042835+0.128505+0.266895+0.3431518+0.205891]=1-0.9872778=0.0127222$

Since, $p(x≤11)=0.0556> 0.05$ then it falls on the Accepted region and further, $p(x≤10)=0.0127<0.05$ then it falls on the Rejected region. We therefore do not reject the Null hypothesis.

Any insight welcome...

 

[andrewkirk]

Having only 73% of survey respondents say they're satisfied is prima facie evidence that Pierre's claim of 90% satisfaction may be exaggerated (overconfident).
But the test shows that, if Pierre is correct, a survey of 15 people has a greater than 5% chance of returning 11 or fewer saying they're satisfied. So we cannot reject the possibility that the low satisfaction score arises purely from random variation, with 95% confidence of our rejection being correct. Since we set ourselves the target of 95% confidence, we do not reject that possibility, which means we do not reject the null hypothesis that 90% of customers are satisfied.
Note it says there is not evidence that Pierre is correct (ie not overconfident). Lack of evidence to disprove proposition P does not constitute evidence for proposition P. Here P is the proposition: "90% of Pierre's customers are satisfied".