Z Test and Standard Error

Question

Before televised debates, a poll of 800 registered voters showed 560 in favor of a particular candidate; after the debates a poll of 600 voters showed 450 in favor of the candidate. A political analyst wants to determine if the debates resulted in an increase in popularity.

Calculate the standard error the analyst will use for the z test

SE = sqrt[ p-hat (1 - p-hat) / n ]

sqrt 0.75 (1-0.75)/ 600 + 0.7 (1-0.7)/ 800

SE = 0.023

Perform the z test and enter the p-value below

??

How would i perform the Z Test if i do not know the observation nor the mean?

Answer 1

Two methods are in use: (a) use $H_0$ and 'pool' all successes and all trials to get $\hat p = 1010/1400$ to compute SE. (b) Estimate $\hat p_1$ and $\hat p_2$ to get two variances, add variances, take sqrt to get SE.

Please consult your textbook or class notes to see which method you are using in your course.

Minitab statistical software gives on a choice. Often there is very little difference between the two standard errors.

Pooled method:

Sample    X    N  Sample p
1       560  800  0.700000
2       450  600  0.750000

Difference = p (1) - p (2)
Estimate for difference:  -0.05
Test for difference = 0 (vs < 0):  Z = -2.07  P-Value = 0.019

Separate variance method:

Sample    X    N  Sample p
1       560  800  0.700000
2       450  600  0.750000

Difference = p (1) - p (2)
Estimate for difference:  -0.05
est for difference = 0 (vs < 0):  Z = -2.09  P-Value = 0.019

Notes: (1) Minitab output edited for relevance in both cases. (Omitting confidence intervals and results of Fisher's Exact test, which gives P-value 0.022.).

(2) Also can use chi-squared test on 2-by-2 table of Yes and No opinions, but that is inherently a two-sided test so the P-value is about doubled. From R:

 Y = c(560, 459)
 All = c(800, 600)
 N = All - Y
 TAB = rbind(Y,N); TAB
    [,1] [,2]
 Y  560  459
 N  240  141

chisq.test(TAB, cor=F)  # Decline Yates' Correction

        Pearson's Chi-squared test

data:  TAB
X-squared = 7.313, df = 1, p-value = 0.006846