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(a):

The probability that after the speech people favor the major is the sum of \({p_1}\)and \({p_3}\)- both are the corresponding probabilities to cells 1 and 3. The probability that before the speech people do not favor the major is the sum of \({p_1}\)and \({p_2}\)both are the corresponding probabilities to cells 1 and 2. This is because the first column represents success for "after" and the first row represents the row for "before" success.

From\({p_1}\)+\({p_3}\)=\({p_1}\)+\({p_2}\), which would be the null hypothesis,

The hypotheses of interest are \({H_0}:{p_3} = {p_2}\) versus \({H_a}:{p_3} > {p_2}\)

(b):

Every cell in the table represents number of people for corresponding described event, this indicates that using that data you could estimate the probabilities. There are total of n individuals, and\({x_1} + {x_2} + {x_3} + {x_4} = n\). From this, the estimate of \({p_1} + {p_3} - \left( {{p_1} + {p_2}} \right)\) is simply

\(\frac{{{x_1} + {x_3} - \left( {{x_1} + {x_2}} \right)}}{n} = \frac{{{x_3} - {x_2}}}{n}\)

W hich is obviously the after/before difference in success probabilities estimate. \(\sigma = \sqrt {\frac{{{p_3} + {p_2}}}{n}} ,\)However, the estimator requires random variables and it is\(\frac{{{X_3} - {X_2}}}{n}\)

(c):

Assume that \({H_0}\)is true, thus\({p_3} = {p_2}\). Based on the given information, the variance of the given estimator in (b) is

\({\mathop{\rm Var}\nolimits} \frac{{{X_3} - {X_2}}}{n} = \frac{{{p_2} + {p_3} - {{\left( {{p_2} - {p_3}} \right)}^2}}}{n}\)

which is, for \({p_3} = {p_2}\)

\({\mathop{\rm Var}\nolimits} \frac{{{X_3} - {X_2}}}{n} = \frac{{{p_3} + {p_2}}}{n}.\)

The standard deviation is which needs to be estimated with\(\hat \sigma = \sqrt {\frac{{{{\hat p}_3} + {{\hat p}_2}}}{n}} \)

Since the expected value is obviously zero, and random variable

\(\frac{{{X_3} - {X_2}}}{n}\)has approximately normal distribution with parameter mean 0 and standard deviation\(\sigma \), you can construct the Z statistic as

\(\begin{array}{c}Z = \frac{{\frac{{{X_3} - {X_2}}}{n}}}{{\sqrt {\frac{{{{\hat p}_3} + {{\hat p}_2}}}{n}} }}\\ = \frac{{{X_3} - {X_2}}}{{\sqrt {{X_3} + {X_2}} }}\end{array}\)

which has standard normal distribution.

(d):

the hypotheses of interest are \({H_0}:{p_3} = {p_2}\) versus\({H_a}:{p_3} > {p_2}\), and the value of the Z test statistic is

\(\begin{array}{c}z = \frac{{{X_3} - {X_2}}}{{\sqrt {{X_3} + {X_2}} }}\\ = \frac{{200 - 150}}{{\sqrt {200 + 150} }}\\ = 2.68\end{array}\)

The test is one sided, upper, thus the P value is the area under the standard normal curve to the right of z. Therefore, the P value is

\(\begin{array}{c}P = P(Z > 2.68)\\ = 1 - P(Z \le 2.68)\\ = 1 - \Phi (2.68)\\ = 0.0037\end{array}\)

where the \(\Phi ( \cdot )\)value can be obtained from the table in the appendix. Depending on significance level, the conclusion might change. At significance level 0.01 or 0.05 which are usually used, because

\(P = 0.0037 < 0.01 < 0.05\)we reject the null hypothesis

However, you might not reject in for other significance levels.