A survey of 542 consumers reveals that 301 favor the new design for a product. Construct a 90% confidence interval for the true proportion of all consumers who favor the design.

The choices are:

0.518 < p < 0.593

0.522 < p < 0.588

0.500 < p < 0.610

0.514 < p < 0.597

0.520 < p < 0.591

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 = 542, \pi = \frac{301}{542} = 0.555[/tex]

90% confidence level

So [tex]\alpha = 0.1[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.555 - 1.645\sqrt{\frac{0.555*0.445}{542}} = 0.52[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.555 + 1.645\sqrt{\frac{0.555*0.445}{542}} = 0.59[/tex]

So the correct answer is:

0.520 < p < 0.591

A survey of 542 consumers reveals that 301 favor the new design for a product. Construct a 90% confidence interval for the true proportion of all consumers who favor the design. The choices are: 0.518 < p < 0.593 0.522 < p < 0.588 0.500 < p < 0.610 0.514 < p < 0.597 0.520 < p < 0.591

Answer :

Other Questions

A survey of 542 consumers reveals that 301 favor the new design for a product. Construct a 90% confidence interval for the true proportion of all consumers who favor the design.

The choices are:

0.518 < p < 0.593

0.522 < p < 0.588

0.500 < p < 0.610

0.514 < p < 0.597

0.520 < p < 0.591