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simpatique

spot speed

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I ran across a problem regarding spot speed ( answer not verified ) . Can you show how to solve it? Thank you

Consider the following two spot speed samples conducted a test location to determine the

effectiveness of a new speed limit posting at 50mph.

Before After

Average Speed 55.3mph 52.8mph

Standard Deviation 5.0mph 5.6mph

Sample Size 100 85

a. Was the new speed limit effective in reducing average speeds at this location?

b. Was the new speed limit effective in reducing average speeds to 50mph?

I tried to use the Z test as I understood it but I am getting Z < 0 so I know I am doing something wrong here .

I'd appreciate your help

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Basic statistics... you need to find the margin of error for each sample. That is, how confident are you that a sample mean, x, is close to the population ("real") mean, μ. A 95% confidence level is pretty standard and easy to remember (Z=1.96)

Margin of Error = Critical Value x Standard Error of the statistic

SEx = σ / sqrt(n)

SEb = 5.0 / sqrt (100) = 0.50

SEa = 5.6 / sqrt (85) = 0.61

For the critical value, use the z-score (aka standard score) that indicates how many standard deviations an element is from the mean:

z = (X - μ) / σ = 1.96 for a 95% confidence level

MEb = 1.96 * 0.50 = 0.98

MEa = 1.96 * 0.61 = 1.19

This relates to confidence interval which is the sample statistic +/- the margin of error.

So... in plain English, you can say this:

You are 95% confident that the real mean (i.e. the population mean) of the speed limit before installing the new speed limit sign was between 54.32 and 56.28 and the real mean of the speed limit after installing the new speed limit sign was between 51.61 and 53.99. And that means you are more than 95% confident the new speed limit sign was effective in reducing average speeds at this location.

I'll let you work through the second question yourself with the following pointer: given the sample mean of 52.8, what is the confidence level that the population mean is really 50.0? If you understand hypothesis testing, this is pretty straight-forward. And you could (should?) have framed this entire problem in the context of hypothesis testing from the beginning.

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I alwyas learn from you Mr IlPADRINO . I learned alot form your blogs , I even dowloaded some of your stuff from wiki engineer .You are a true asset to this forum

Basic statistics... you need to find the ma rgin of error for each sample. That is, how confident are you that a sample mean, x, is close to the population ("real") mean, μ. A 95% confidence level is pretty standard and easy to remember (Z=1.96)

Margin of Error = Critical Value x Standard Error of the statistic

SEx = σ / sqrt(n)

SEb = 5.0 / sqrt (100) = 0.50

SEa = 5.6 / sqrt (85) = 0.61

For the critical value, use the z-score (aka standard score) that indicates how many standard deviations an element is from the mean:

z = (X - μ) / σ = 1.96 for a 95% confidence level

MEb = 1.96 * 0.50 = 0.98

MEa = 1.96 * 0.61 = 1.19

This relates to confidence interval which is the sample statistic +/- the margin of error.

So... in plain English, you can say this:

You are 95% confident that the real mean (i.e. the population mean) of the speed limit before installing the new speed limit sign was between 54.32 and 56.28 and the real mean of the speed limit after installing the new speed limit sign was between 51.61 and 53.99. And that means you are more than 95% confident the new speed limit sign was effective in reducing average speeds at this location.

I'll let you work through the second question yourself with the following pointer: given the sample mean of 52.8, what is the confidence level that the population mean is really 50.0? If you understand hypothesis testing, this is pretty straight-forward. And you could (should?) have framed this entire problem in the context of hypothesis testing from the beginning.

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I alwyas learn from you Mr IlPADRINO . I learned alot form your blogs , I even dowloaded some of your stuff from wiki engineer .You are a true asset to this forum

Basic statistics... you need to find the ma rgin of error for each sample. That is, how confident are you that a sample mean, x, is close to the population ("real") mean, μ. A 95% confidence level is pretty standard and easy to remember (Z=1.96)

Margin of Error = Critical Value x Standard Error of the statistic

SEx = σ / sqrt(n)

SEb = 5.0 / sqrt (100) = 0.50

SEa = 5.6 / sqrt (85) = 0.61

For the critical value, use the z-score (aka standard score) that indicates how many standard deviations an element is from the mean:

z = (X - μ) / σ = 1.96 for a 95% confidence level

MEb = 1.96 * 0.50 = 0.98

MEa = 1.96 * 0.61 = 1.19

This relates to confidence interval which is the sample statistic +/- the margin of error.

So... in plain English, you can say this:

You are 95% confident that the real mean (i.e. the population mean) of the speed limit before installing the new speed limit sign was between 54.32 and 56.28 and the real mean of the speed limit after installing the new speed limit sign was between 51.61 and 53.99. And that means you are more than 95% confident the new speed limit sign was effective in reducing average speeds at this location.

I'll let you work through the second question yourself with the following pointer: given the sample mean of 52.8, what is the confidence level that the population mean is really 50.0? If you understand hypothesis testing, this is pretty straight-forward. And you could (should?) have framed this entire problem in the context of hypothesis testing from the beginning.

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I alwyas learn from you Mr IlPADRINO . I learned alot form your blogs , I even dowloaded some of your stuff from wiki engineer .You are a true asset to this forum

Basic statistics... you need to find the ma rgin of error for each sample. That is, how confident are you that a sample mean, x, is close to the population ("real") mean, μ. A 95% confidence level is pretty standard and easy to remember (Z=1.96)

Margin of Error = Critical Value x Standard Error of the statistic

SEx = σ / sqrt(n)

SEb = 5.0 / sqrt (100) = 0.50

SEa = 5.6 / sqrt (85) = 0.61

For the critical value, use the z-score (aka standard score) that indicates how many standard deviations an element is from the mean:

z = (X - μ) / σ = 1.96 for a 95% confidence level

MEb = 1.96 * 0.50 = 0.98

MEa = 1.96 * 0.61 = 1.19

This relates to confidence interval which is the sample statistic +/- the margin of error.

So... in plain English, you can say this:

You are 95% confident that the real mean (i.e. the population mean) of the speed limit before installing the new speed limit sign was between 54.32 and 56.28 and the real mean of the speed limit after installing the new speed limit sign was between 51.61 and 53.99. And that means you are more than 95% confident the new speed limit sign was effective in reducing average speeds at this location.

I'll let you work through the second question yourself with the following pointer: given the sample mean of 52.8, what is the confidence level that the population mean is really 50.0? If you understand hypothesis testing, this is pretty straight-forward. And you could (should?) have framed this entire problem in the context of hypothesis testing from the beginning.

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space M speed is 2%less than timeMspeed the new time mean speed is 52.8 therefore the new space mean speed should be 51.74 . with 85 observations 95 % confidence = 1.9599(standard error) = 1.1975

95 % confi spot speed = 51.74+- 1.1975 50 mph speed limit should be 85 % of 52.93 or 50.5425 but 85 % of 52.93 = 44.99 mph < 50 mph.. No go

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space M speed is 2%less than timeMspeed the new time mean speed is 52.8 therefore the new space mean speed should be 51.74 . with 85 observations 95 % confidence = 1.9599(standard error) = 1.1975

95 % confi spot speed = 51.74+- 1.1975 50 mph speed limit should be 85 % of 52.93 or 50.5425 but 85 % of 52.93 = 44.99 mph < 50 mph.. No go

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space M speed is 2%less than timeMspeed the new time mean speed is 52.8 therefore the new space mean speed should be 51.74 . with 85 observations 95 % confidence = 1.9599(standard error) = 1.1975

95 % confi spot speed = 51.74+- 1.1975 50 mph speed limit should be 85 % of 52.93 or 50.5425 but 85 % of 52.93 = 44.99 mph < 50 mph.. No go

I don't follow.

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Av speed = mean speed +or - 1.96 ( standard deviation) speed limit is 85% of average speed ...= space mean speed +- 1.96 (satandard deviation )

On average space mean speed is 2% less than time mean speed

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Anyway this problem is a goood problem

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