Question

Answer all of the following parts of the question(s)

(Part A)

In a study regarding public opinion about ObamaCare, how many people (at a minimum) should be included in a sample to be 95% sure that the sample estimate is within three percentage points of the population proportion p?

Group of answer choices

(A) 1068

(B) 1067

(C) 752

(D) 33

(Part B)

Specify the rejection region associated with the test of H_0: mu = 10, H_a: mu > 10 when alpha = 0.05, and n = 17 and sample is drawn from normal distribution.

Group of answer choices

(A) t > 2.12

(B) Z > 1.645

(C) t > 1.746

(D) Z < 1.96

(Part C)

In a test of H_0: mu = 100 against H_a: mu < > 100, the sample data yielded the test statistic z = 2.17. Find the p-value for the test. Here "<>" stands for "not equal".

Group of answer choices

(A )0.485

(B )0.03

(C) 0.015

Answer 1

(Part A)

For sample estimate to be within 3 points, the Margin of Error of the confidence interval should be 3

Margin of error is given by

$z\times\sqrt{\frac{\hat{p} (1-\hat{p})}{n}}$

Since sample proportion is not given, we assume $\hat{p}=0.5$

For 95% Confidence interval
α = 0.05,      α/2 = 0.025
From z tables of Excel function NORM.S.INV(α/2) we find the z value
z = NORM.S.INV(0.025) = 1.96
We take the positive value of z

Thus

$\newline z\times\sqrt{\frac{\hat{p} (1-\hat{p})}{n}} = 0.03 \newline 1.96\times\sqrt{\frac{0.5 (1-0.5)}{n}} = 0.03 \newline \newline n = 1067.111 \newline n = 1068$... rounding up

Answer :

(A)     1068

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(Part B)

The null and alternative hypotheses are

H_0: mu = 10,

H_a: mu > 10

alpha = 0.05, and n = 17

Since the population standard deviation is unknown, and sample size is small, we use t-distribution

This is a right tailed test as the alternate hypothesis contains greater than sign

Degrees of freedom = 17 - 1 = 16

Using alpha, degrees of freedom and Excel function t.inv we find the critical value of t

t-critical = -(T.INV(0.05, 16))                              (Negative sign is because it is a right tailed test)

t-critical = 1.746
Rejection region : Reject Ho when t > 1.746

Answer :

(C)       t > 1.746

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(Part C)

H_0: mu = 100 against H_a: mu < > 100

z = 2.17

Since the alternate hypothesis contains "not equal to" it is a two sided test

We calculate p-value using Excel function NORM.S.DIST

p-value = 2*(1-NORM.S.DIST(2.17, TRUE))                      (We first find the left side area and then multiply by 2

                                                                                              as it is a 2 sided test)

p-value = 0.03

Answer :

(B)       0.03

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