A tailor thinks that about 75 % of customers are satisfied with service. A random sample of 15 customers, 8 said they were satisfied. Calculate whether this is plausible at 5 % significance level.I figured that μ = 0.6, and x ¯ = 8 / 15, but I'm not sure how to get the variance. Can you assume satisfaction follows a Bernoulli distribution and calculate variance as ( 0.75 ( 1 − 0.75 ) ) / n?

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A tailor thinks that about   75  % of customers are satisfied with service. A random sample of   15 customers,   8 said they were satisfied. Calculate whether this is plausible at   5  % significance level.I figured that   μ  =  0.6, and       x    ¯    =  8  /  15, but I&#039;m not sure how to get the variance. Can you assume satisfaction follows a Bernoulli distribution and calculate variance as   (  0.75  (  1  −  0.75  )  )  /  n?

Cahokiavv · Accepted Answer

With such a small sample, a normal approximation is inappropriate. Instead, the exact binomial probability is simultaneously easier to understand and guarantees the correct Type I error control.Consider a rephrasing of the question as follows. You sample   n  =  15 customers, and you observed   X  =  8 satisfied customers. Under the null hypothesis that the true probability of customer satisfaction is   p  =  0.75, what is the probability of obtaining a result at least as extreme as the one observed,   X  =  8?To compute this, you first compute  Pr  [  X  =  8  ∣  p  =  0.75  ]  =                (              15      8                  )        (  0.75      )    8    (  0.25      )    7    ≈  0.0393205.Now, among all the other possible outcomes   X  ∈  {  0  ,  1  ,  2  ,  …  ,  15  }, which of these will have lower probability? Since   8  /  15  &amp;lt;  0.75, we know that   X  ∈  {  0  ,  1  ,  2  ,  …  ,  7  } will each have smaller probability than   Pr  [  X  =  8  ]. But what about the upper tail? The best way is to compute from   X  =  15 down. We have  Pr  [  X  =  15  ∣  p  =  0.75  ]  ≈  0.0133635    Pr  [  X  =  14  ∣  p  =  0.75  ]  ≈  0.0668173and we stop here because this probability already exceeds   Pr  [  X  =  8  ∣  p  =  0.75  ]. So the exact   p-value is                    p                            =        Pr        [        X        ≤        8        ∣        p        =        0.75        ]        +        Pr        [        X        =        15        ∣        p        =        0.75        ]                                          ≈        0.0566203        +        0.0133635                                          =        0.0699838.            This exceeds the significance level   α  =  0.05, so the observed data does not furnish enough evidence to reject the hypothesis that   p  =  0.75 at the   5  % significance level.Note that various scientific calculators can compute binomial CDFs, or you can compute them manually; e.g.  Pr  [  X  ≤  8  ∣  p  =  0.75  ]  =      ∑          x      =      0        8                  (              15      x                  )        (  0.75      )    x    (  0.25      )          15      −      x        ,

A tailor thinks that about 75 <mi mathvariant="normal">% of customers are

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