Test Hypothesis

Question

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

Too many questions and no effort. What do you expect ? – callculus42 Dec 03 '15 at 01:06 — callculus42, Dec 03 '15 at 01:06
There was an effort, help me start. – Bhavik P. Dec 03 '15 at 01:08 — Bhavik P., Dec 03 '15 at 01:08

callculus42 · Accepted Answer · 2015-12-03T06:27:07.060

0

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.

edited Dec 03 '15 at 06:27

answered Dec 03 '15 at 05:59

callculus42

30,550

This makes much more sense. I kept getting confused with my t-value. I was also using n to be 15 instead of using the sample size. Thank you callculus. I appreciate the help. – Bhavik P. Dec 03 '15 at 06:20
@BhavikP. You are welcome. I´m sure that you will manage to draw the graph with the two tails. – callculus42 Dec 03 '15 at 06:25
yes, I think I got that part under control – Bhavik P. Dec 03 '15 at 06:28

score -2 · Answer 2 · edited Dec 03 '15 at 02:33

-2

its a p test from what I understood so test statistic is from $z=(p'-p)/((pq/n)^{.5})$

edited Dec 03 '15 at 02:33

Ben Sheller

4,085

answered Dec 03 '15 at 02:05

user295417

1

Please look at the edit of the work I have done. Does it look accurate? – Bhavik P. Dec 03 '15 at 03:22

Test Hypothesis

2 Answers2