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[SOLVED] An urban highway has a speed limit of 50 km/ h

karush

Well-known member
Jan 31, 2012
2,740
IBP9 An urban highway has a speed limit of 50 km/ h

View attachment 1217

(a) assume the $km\ h^{-1}$ means $\displaystyle\frac{km}{h}$

$30\%$ on Z-table is $\approx 0.53$

so using $\displaystyle\frac{X-\mu}{\sigma}=Z$ then $\displaystyle\frac{50-44.8}{10}=0.53$

(b) (c) (d) to come
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You want to find the $z$-score associated with an area of 0.2, because you want 70% of the data to the left of your $x$ value and 30% to the right. Using a table and interpolating a bit, we find this $z$-score is about 0.525 (Okay, we have different style tables; when I saw you approaching your table with 30%, I thought you were not understanding correctly what to do). :D

W|A returns a $z$-score of $z\approx0.524401$.

We know the data value associated with this $z$-score is 50 (the speed limit), and so to find the mean:

\(\displaystyle \mu=x-z\sigma\approx50-5.24401=44.75599\approx44.8\)
 

karush

Well-known member
Jan 31, 2012
2,740
W|A returns a $z$-score of $z\approx0.524401$.

\(\displaystyle \mu=x-z\sigma\approx50-5.24401=44.75599\approx44.8\)
not sure how to use W|A to get this

erf(z/sqrt(2))=2(0.4) was a previous example but is the input for this problem?

also (b) (c) and (d) somewhat clueless on these.:confused:
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
not sure how to use W|A to get this

erf(z/sqrt(2))=2(0.4) was a previous example but is the input for this problem?

also (b) (c) and (d) somewhat clueless on these.:confused:
In this case use:

erf(z/sqrt(2))=2(0.2)

It's been 20 years since I took statistics, and to be honest, I have forgotten about the null hypothesis, but I'm sure someone here is more knowledgeable about this. :D
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,793
Re: IBP9 An urban highway has a speed limit of 50 km/ h

View attachment 1217

(a) assume the $km\ h^{-1}$ means $\displaystyle\frac{km}{h}$

$30\%$ on Z-table is $\approx 0.53$

so using $\displaystyle\frac{X-\mu}{\sigma}=Z$ then $\displaystyle\frac{50-44.8}{10}=0.53$

(b) (c) (d) to come
I'll just show the method in this case.

(b) $H_1$: the mean speed after the campaign is less than $\mu_0 = 44.8\text{ km/h}$.

(c) This is one-tailed because the police is only interested in "less than".

(d) The corresponding test statistic is:
$$z = \frac {\bar x - \mu_0} {\sigma / \sqrt n} = \frac {41.3 - 44.8} {10 / \sqrt{25}}$$

If this z-score is less than the critical z-score corresponding to 5% probability, we can conclude that the campaign has been effective.

Can you finish (d)?
 

karush

Well-known member
Jan 31, 2012
2,740
I'll try

from $\displaystyle z = \frac {\bar x - \mu_0} {\sigma / \sqrt n} =
\frac {41.3 - 44.8} {10 / \sqrt{25}}=-1.75$

which from $0$ to $Z$ is $4\%$ this is less than the $5\%$ required for an effective campaign
so it was effective.(Wasntme)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,793
Good! :)