Question

True or False: A sample of 100 fuses from a very large shipment is found to have 10 that are defective. The 0.95 confidence interval would indicate that, for this shipment, the proportion of defective fuses is between 0 and 0.28. False True

Answer 1

`0.10 - 1.96\sqrt{(0.1(1-0.1))/(100)}=0.0412`

`0.10 + 1.96\sqrt{(0.1(1-0.1))/(100)}=0.1588`

The 95% confidence interval for the proportion of defectives would be given by (0.0412;0.1588)

So then we can conclude that the statement given is FALSE.

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The estimated proportion of defectives for this case is
`\hat p = (10)/(100)=0.1`

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by
`\alpha=1-0.95=0.05` and
`\alpha/2 =0.025`. And the critical value would be given by:

`z_(\alpha/2)=-1.96, z_(1-\alpha/2)=1.96`

The confidence interval for the mean is given by the following formula:

`\hat p \pm z_(\alpha/2)\sqrt{(\hat p (1-\hat p))/(n)}`

If we replace the values obtained we got:

`0.10 - 1.96\sqrt{(0.1(1-0.1))/(100)}=0.0412`

`0.10 + 1.96\sqrt{(0.1(1-0.1))/(100)}=0.1588`

The 95% confidence interval for the proportion of defectives would be given by (0.0412;0.1588)

So then we can conclude that the statement given is FALSE.