Question

Police recorded the average speed of cars driving on a busy street by a school. For a sample of 36 ​speeds, it was determined that the average amount over the speed limit for the 36 speeds was 13.2 mph with a standard deviation of 8 mph. The 90​% confidence interval estimate for this sample is 10.95 mph to 15.45 mph. ​

a) What is the margin of error for this​ problem? ​
b) What size sample is needed to reduce the margin of error to no more than plus or minus1​?

Answer 1

a)
`\pm 2.2527`

b) n = 169

Explanation:

We are given the following information:

Sample mean,
`\bar{x}` = 13.2 mp

Sample size, n = 36

Alpha, α = 0.10

Sample standard deviation, s = 8 mph

90% Confidence interval: (10.95, 15.45)

a)

`\text{Margin of error}} = \text{Critical value}* \text{Standard error of the statistic}\\ = t_(critical)* \displaystyle(s)/(√(n))`

`t_(critical)\text{ at degree of freedom 35 and}~\alpha_(0.10) = \pm 1.689572`

Putting the values, we get,

`\text{Margin of error} = t_(critical) * \displaystyle(s)/(√(n))\\\\= \pm 1.689572* (8)/(√(36)) = \pm 2.2527`

b)
`\text{Margin of error} = \pm 1`

`t_(critical) * \displaystyle(s)/(√(n))\\\\\Rightarrow \pm 1.689572* (8)/(√(n)) = \pm 1\\\\\Rightarrow √(n) = (\pm 1.689572* 8)/( \pm 1) = 13.5165 \approx 13\\\\\Rightarrow \bold{n = 169}`