Hypothesis Test on Bank Service Ratings

[toothpaste666]

Homework Statement

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.

The Attempt at a Solution

a) the proportion of customers in the sample who rated the service as excellent is
258/505
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?

 

[RUber]

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.

 

[toothpaste666]

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