Question

2) In the last problem set, we considered a survey of business owners about parking availability, with N-100 respondents. 56 of these respondents said that a shortage of parking is a constraint on their business's profitability. We are still interested in whether we can be confident that more than 50% of owners see parking shortages as a problem, but now we have more tools to bring to bear. a) Test the above hypothesis, using a 90% confidence level. Do you reject the hypothesis? b) Did you use a one-side or a two-sided test here? c) Explain in words what your answer to (a) means d) What is the p-value for the hypothesis? e) Explain in words what your answer to (d) means.

Answer 1

a)

Let p be the true proportion of owners who see parking shortages as a problem. The appropriate hypotheses are,

H0: p = 0.5

Ha: p > 0.5

If we assume the null hypothesis is true, standard error of sample proportion is,

SE = = 0.05

p(1 - p)/N = 0.5(1 – 0.5)/100

Z value for 90% confidence level is 1.645

Margin of error = z * SE = 1.645 * 0.05 = 0.08

Sample proportion, $\hat{p}$ = 56/100 = 0.56

90% confidence level is,

(0.56 - 0.08, 0.56 + 0.08)

(0.48, 0.64)

Since the 90% confidence level contains the hypothesized value of proportion, we fail to reject H0, and conclude that there is no significant evidence that true proportion of owners who see parking shortages as a problem is greater than 0.5

b)

Since the alternative hypothesis tests for p > 0.5, it is a one-sided test.

c)

The given data does not provide sufficient evidence that the true proportion of owners who see parking shortages as a problem is greater than 0.5.

d)

Test statistic, Z = ($\hat{p}$ - p) / SE = (0.56 - 0.5) / 0.05 = 1.2

P-value = P(z > 1.2) = 0.1151

e)

The probability of finding the observed, or more extreme, results is 0.1151 when the null hypothesis (H0: p = 0.5) is true.