Question

2. In the modern political era, campaign rallies are being infiltrated by people who do not sup- port the speaker (e.g., to sow dissent, to study those who do support the speaker, etc.). You're curious about this, so you decide to attend one of President Trump's rallies. Your plan is sim ple: you'll choose 70 random people in the audience and use hidden cameras to videotape them during the rally. When the crowd breaks into its third chant (e.g., "Lock her up!" or "Build the wal), you wil use the video footage to see what proportion of your random subjects actually engage in the chant. Suppose you do this and find only 58 of the 70 people took part in the chant. a. Assign notation to and define the population parameter we are trying to study. Then, create an approximate 92% CI for this parameter b. You may have noticed that your approximate 92% CI from part a did not include 100% (or 1, if you are using decimals). Suppose you change the confidence level to C% and the upper bound of the approximate CI exactly equals 100%. Find C.

Answer 1

A) $\widehat p$ = 58/70 = 0.83

At 92% confidence interval the critical value is z_0.04 = 1.75

The 92% confidence interval for population proportion is

$\widehat p$ +/- z_0.04 * sqrt($\widehat p$(1 - $\widehat p$)/n)

= 0.83 +/- 1.75 * sqrt(0.83 * (1 - 0.83)/70)

= 0.83 +/- 0.08

= 0.75, 0.91

B) 0.83 + z^* * sqrt(0.83 * (1 - 0.83)/70) = 1

Or, z^* * sqrt(0.83 * 0.17/70) = 1 - 0.83

Or, z^* = 0.17/(sqrt(0.83 * 0.17/70)

Or, z^* = 3.786

At 100% confidence level the critical value z^* will be 3.786.

So the confidence level is 100%.