SUV vs Minivan propotion

[Timcago]

A sample survey is designed to estimate the proportion of sports utility vehicles and minivans being driven in the state of California. A random sample of 500 registrations is selected from a DMV database. 68 are classified as sports utility vehicles. A second random sample of 600 registrations is selected, and 84 are classified as minivans. Is there evidence to suggest a difference in the proportion of sports utility vehicles and the proportion of minivans in California, with a significance level of .0025? Justify your answer.

SuV:

p=68/500 = .136

Zc= 1.96?

E=1.96([sqrt]((.136)(.864))/(500))= .030047

.136-.030047<P<.136+.030047

.106<P<.166

Minivan:

p=84/600 =.14

Zc=1.96?

E=1.96([sqrt]((.14)(.86))/(600))= .0278

.14-.0278<P<.14+.0278

.112<P<.168

Since the intervals overlap there is not enough evidence to indicate a difference in the proportion of SUVs and minivans in California.


Is this done correctly? I have done problems similar to this in the past except they came from the same sample. This one has 2 separate samples so i was wondering if i needed to do it differently. The significance thing is new to me aswell so i am hoping i got my z-score correct.

Can someone check this?

 

[galactus]

Test stat = -0.1914
Z value = +/- 3.0234
P value = 0.8482

Two-tail test

Do not reject.

There is not enough evidence to support the claim that there is a difference.

 

[Timcago]

Can you walk me through how you got these numbers?

I am very confused.

How did you get the test stat, the Z value, and the P value?

 

[galactus]

Actually, I have a wonderful Excel macro that does about every stats thing I need. I ran it through that. Besides, most any TI will do this stuff.
Who does it by hand anymore?.

But if you must.

\(\displaystyle n_{1}=500, \;\ p_{1}=0.136=\frac{17}{125}, \;\ x_{1}=68, \;\ n_{2}=600, \;\ p_{2}=0.14=\frac{7}{50}, \;\ x_{2}=84\)

\(\displaystyle \L\\\overline{p}=\frac{x_{1}+x_{2}}{n_{1}+n_{2}}=\frac{68+84}{500+600}=\frac{38}{275}=0.13\overline{81}\)

\(\displaystyle \L\\\overline{q}=1-\overline{p}=\frac{237}{275}\)

To find the test statistic:

\(\displaystyle \L\\z=\frac{(p_{1}-p_{2})}{\sqrt{\overline{p}\overline{q}\left(\frac{1}{n_{1}}+\frac{1}{n_{2}}\right)}}\approx\frac{\frac{17}{125}-\frac{7}{50}}{\sqrt{\frac{38}{275}\cdot\frac{237}{275}\left(\frac{1}{500}+\frac{1}{600}\right)}}=-0.1914\)

The critical value can be found from your 0.0025. Look up what z-score corresponds to 0.99875 in the body of the table.


PHStat2, Excel Add-In:

Data
Hypothesized Difference 0
Level of Significance 0.0025
Group 1
Number of Successes 68
Sample Size 500
Group 2
Number of Successes 84
Sample Size 600

Intermediate Calculations
Group 1 Proportion 0.136
Group 2 Proportion 0.14
Difference in Two Proportions -0.004
Average Proportion 0.138181818
Z Test Statistic -0.191421625

Two-Tail Test
Lower Critical Value -3.023358193
Upper Critical Value 3.023358193
p-Value 0.848195311
Do not reject the null hypothesis