Test Hypothesis I need help solving the following problem. I exhausted many options however I fell short. Any help is appreciated.
Creative Good, a New York consulting firm, claimed that 35% of shoppers fail in their attempt to purchase merchandise on-line because Web sites are too complex. A random sample of 60 on-line shoppers was selected and 15 reported sufficient frustration with their sites to deter making a purchase.  Do these data provide sufficient evidence that percentage of shoppers fail in their attempt to purchase on-line differs from 35% ?
Mainly need help with the bolded questions!
a. Set up the null and alternate hypothesis
b. Calculate the test statistic.
c. Draw the curve and find the critical value(s).
Work done so far
 A: The hypotheses and the test statistic, $T=-1.624$, are right. Your t-value is not right. The parameter $n$ at $t_{n-1,\frac{\alpha}{2}}$ and $t_{n-1,1-\frac{\alpha}{2}}$ is the sample size. Therefore $n-1=59$. Now it depends what significance level you decide to take. Let´s say it is $\alpha=10\%=0.1$. Thus $\frac{\alpha}{2}=0.05$ and $ 1-\frac{\alpha}{2}=1-0.05=0.95$.
$t_{59,0.05}=-1.671$
$t_{59,0.95}=1.671$
Therefore T is inside the 90% interval, because $-1.671<-1.624<1.671$
$H_0$ can not be rejected at the significance level of $10\%$. You have the two critical values. That means that you have two areas (left and right) which represents the lower 5% and the upper 5% of the t-distribution.
Remark
Because of the thumb rule $n\cdot p \cdot (1-p)>9$ the t-distribution can be approximated by the normal distribution.
$z_{0.05}=-1.645$ and $z_{0.95}=1.645$. The test statistic is inside the confidence interval again.
A: its a p test from what I understood so test statistic is from $z=(p'-p)/((pq/n)^{.5})$
