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

[karush]

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

 

[MarkFL]

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).

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]

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.

 

[MarkFL]

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.

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.

 

[Klaas van Aarsen]

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]

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.

 

[Klaas van Aarsen]

Good!