Confidence Intervals, Edexcel S3Watch

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Thread starter 8 years ago
#1
I am stuck on this question regarding confidence intervals, in particular part b.

The managing director of a certian firm has commisioned a survey to estimate the mean expenditure of customers on electrical appliances. A random sample of 100 people were questioned and the research team presented the managing director with a 95% confidence interval of (£128.14, £141.86).

The director says that this interval is too wide and wants a confidence interval of total width £10.

a) Using the same value of , find the confidence limits in this case.

b) Find the level of confidence for the interval in part (a).

Ok, so, I can do part (a) but part (b) is where it gets difficult.

Any help would be greatly appreciated.

Thanks.
0
8 years ago
#2
xbar + or - z*sigma/rootN gives you the boundaries of the Confidence interval.

The width is 2z*sigma/rootN, put this equal to 10 then solve for z. Use the normal distribution tables to figure out the level of confidence.
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Thread starter 8 years ago
#3
(Original post by vc94)
xbar + or - z*sigma/rootN gives you the boundaries of the Confidence interval.

The width is 2z*sigma/rootN, put this equal to 10 then solve for z. Use the normal distribution tables to figure out the level of confidence.
I tried what you suggested, and you get 2 of the same equation -

z*sigma/rootN = 5

On the CD where the solutions are, it has that equation but it also has another equation -

1.96*sigma/rootN = 6.86

Where does this come from?

btw, * = multiply
0
8 years ago
#4
Ok, for part (a) you got xbar = 135 (midpoint of interval).
The given interval has width 13.72, so

2*1.96*sigma/rootN = 13.72, where the 1.96 is because of the 95% CI.
So 1.96*sigma/rootN = 6.86, solving gives sigma=35.

So for part (b), put the width 2*z*sigma/rootN = 10 to get a new z-value and hence the % level of confidence!
0
3 years ago
#5
I'm having problems for part b, can anyone clarify it for me?
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