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Homework Help: A hypothesis test

  1. Nov 21, 2015 #1
    1. The problem statement, all variables and given/known data

    An independent bank, concerned about its customer base, decided to conduct a survey of bank customers. Out of 505 customers who returned the survey form, 258 rated the overall bank services as excellent.

    (a) Test, at level α = .10, the null hypothesis that the proportion of customers who would rate the overall bank services as excellent is .46 versus a two-sided alternative

    (b) Calculate the p-value and comment on the strength of evidence.

    3. The attempt at a solution
    a) the proportion of customers in the sample who rated the service as excellent is
    the null hypothesis is that μ = .46. The alternatives are that μ < .46 or μ > .46. we reject the null when Z> 1.645 or Z < -1.645

    Z = (X-μ)sqrt(n)/s = (258/505 - .46)sqrt(505)/s

    but I am running into a problem because I don't know s so I am wondering if I did this wrong?
  2. jcsd
  3. Nov 23, 2015 #2


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    Homework Helper

    s is based on your assumption about the proportions.
    Normally, the variance for proportions is (p)(1-p).
    So the corresponding s is ##\sqrt{p(1-p)}##.
    And from my understanding, you would want to use the p from your null hypothesis rather than the p from your sample.
  4. Nov 25, 2015 #3
    ok I was looking at the wrong chapter of my book. It is a large sample so we use the statistic
    H0 : p = p0
    H1 : p ≠ p0
    Z = (x-np0)/sqrt(np0(1-p0)) = (258 - 505(.46))/sqrt(505(.46)(.54)) = 2.29
    we reject H0 if Z > zα/2 or Z < -zα/2
    zα/2 = z.1/2= z.05 = 1.645
    Z > zα/2 so we reject the null hypotheses. They are not equal.

    for part b)
    P(Z>2.29) = 1 - F(2.29) = 1 - .9890 = .011
    since it is a two sided test, the p value is twice this
    the p value is .022
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