Respuesta :
Answer:
a) [tex]0.0003095 - 1.64\sqrt{\frac{0.0003095(1-0.0003095)}{420090}}=0.000265[/tex]
[tex]0.0003095 - 1.64\sqrt{\frac{0.0003095(1-0.0003095)}{420090}}=0.000354[/tex]
The 90% confidence interval would be given by (0.000265;0.000354) and in % would be (0.0265, 0.0354)
b) For this case the rate given is 0.0212% and since our interval not contains this value we can conclude at 10% of significance that the real parameter is different from the value given.
Step-by-step 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
Part a
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 90% of confidence, our significance level would be given by [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2 =0.05[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64[/tex]
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
the estimated proportion for this case is:
[tex]\hat p = \frac{130}{420090}=0.0003095[/tex]
If we replace the values obtained we got:
[tex]0.0003095 - 1.64\sqrt{\frac{0.0003095(1-0.0003095)}{420090}}=0.000265[/tex]
[tex]0.0003095 - 1.64\sqrt{\frac{0.0003095(1-0.0003095)}{420090}}=0.000354[/tex]
The 90% confidence interval would be given by (0.000265;0.000354) and in % would be (0.0265, 0.0354)
Part b
For this case the rate given is 0.0212% and since our interval not contains this value we can conclude at 10% of significance that the real parameter is different from the value given.
