A sample of 300 FM residents was taken and 135 said their favorite season was summer. Calculate a 99% confidence interval for the proportion of FM residents whose favorite season is summer. What is the lower bound of this interval? (Round your answer to 3 decimal places and enter it with a leading 0. ex: 0.XXX).

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

[tex]n = 300, \pi = \frac{135}{300} = 0.45[/tex]

99% confidence level

So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.45 - 2.575\sqrt{\frac{0.45*0.55}{300}} = 0.376[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.45 + 2.575\sqrt{\frac{0.45*0.55}{300}} = 0.524[/tex]

The 99% confidence interval for the proportion of FM residents whose favorite season is summer is between (0.376, 0.524).

The lower bound of this interval is 0.376.

dfbustos dfbustos · Answer 2 · 2020-04-15T05:52:55+02:00

Answer:

[tex]0.45 - 2.58\sqrt{\frac{0.45(1-0.45)}{300}}=0.376[/tex]

[tex]0.45 + 2.58\sqrt{\frac{0.45(1-0.45)}{300}}=0.524[/tex]

The 99% confidence interval would be given by (0.376;0.524)

So then the lower bound would be 0.376

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

For this case the estimated proportion of interest would be [tex] \hat p = \frac{135}{300}= 0.45[/tex]

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 99% of confidence, our significance level would be given by [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2 =0.005[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58[/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]

If we replace the values obtained we got:

[tex]0.45 - 2.58\sqrt{\frac{0.45(1-0.45)}{300}}=0.376[/tex]

[tex]0.45 + 2.58\sqrt{\frac{0.45(1-0.45)}{300}}=0.524[/tex]

The 99% confidence interval would be given by (0.376;0.524)

So then the lower bound would be 0.376