Question

12-E3.  From a survey of 557 people, 378 said they preferred your product to your leading competitor. On a repeat survey of 335 people taken a week later, 241 said they prefer your product.

1. Based on the first survey, estimate the proportion of the underlying population who prefer your product. Give a standard error and a 95% confidence interval. Give your answer in terms that would be understood by someone not knowing any statistics.
2. Are the two surveys in agreement statistically? What would it mean if they were not?

Answer 1

(a)

(i)

n = Sample Size = 557

p = Sample Proportion = 378/557 = 0.6786

q = 1 - p = 0.3214

SE =

So,

Standard Error = 0.0198

(ii)

= 0.05

ndf = n - 1 = 557 - 1 = 556

From Table, critical values of t = 1.9642

Confidence Interval:

0.6786 (1.9642 X 0.0198)

= 0.6786 0.0389

= (0.6397    , 0.7175)

Confidence Interval:

0.6397 < P < 0.7175

(iii)

95% confidence interval (0.6397    , 0.7175) is a range of values that is likely to contain unknown population proportion. If the experimentation is repeated many times and confidence interval is calculated each time, 95% of the intervals will contain the population proportion.

(b)

(i)

n1 = 557

p1 = 378/557 = 0.6786

n2 = 335

p2 = 241/335 = 0.7194

Q = 1 - P = 0.3061

Test statistic is given by:

Z = (0.6786 - 0.7194)/0.0319

= - 1.2790

= 0.05

From Table,critical values of Z= 1.96

Since calculated value of Z = - 1.2790 is less than critical value of Z = - 1.96, the difference is significant. Reject null hypothesis.

Conclusion:
The data do not support the claim that the two surveys are in agreement statistically.

(ii)

The two surveys are not statistically not in agreement means that within the given week, the preference of your product to your leading competitor has changed.