A study examined the effectiveness of an ear-molding technique used to correct ear deformities in newborns without the need for surgery. A plastic mold was placed around the deformed ear shortly after birth for about two weeks. The study reports that 152 of 158 procedures were successful. We would like to estimate with 95% confidence the true proportion of successful ear-molding procedures among all newborns needing ear-deformity correction. We can use

Answer:

The 95% confidence interval for the true proportion of successful ear-molding procedures among all newborns needing ear-deformity correction is (0.932, 0.992).

Step-by-step explanation:

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].

The study reports that 152 of 158 procedures were successful.

This means that [tex]n = 158, \pi = \frac{152}{158} = 0.962[/tex]

95% confidence level

So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.962 - 1.96\sqrt{\frac{0.962*0.038}{150}} = 0.932[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.962 + 1.96\sqrt{\frac{0.962*0.038}{150}} = 0.992[/tex]

The 95% confidence interval for the true proportion of successful ear-molding procedures among all newborns needing ear-deformity correction is (0.932, 0.992).

The 95% confidence interval of the true proportion of successful ear-molding procedures among all newborns needing ear-deformity correction is; CI = (0.9322, 0.9918)

We want to find confidence interval and the formula is;

CI = p^ ± z√(p^(1 - p^)/n)

We are told that the study reports that 152 of 158 procedures were successful. Thus;

n = 158

p^ = 152/158

p^ = 0.962

Now, we want to find the confidence interval at a 95% confidence level.

The critical z value for 95% confidence level is;

z = 1.96

Thus;

CI = 0.962 ± 1.96√(0.962(1 - 0.962)/158)

CI = 0.962 ± 0.0298

CI = (0.962 - 0.0298), (0.962 + 0.0298)

CI = (0.9322, 0.9918)

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