Question

3) The approval rating for the superintendent of Nazareth School District was 55% at the beginning of the year. Due to recent events, the superintendent believes his approval rating is increasing. He asks his staff to do a poll. The staff randomly selects 500 adults who have children in the school system. The staff find that 285 of those surveyed approve of the superintendent’s performance. Use a significance level of ? = 0.05.

a) State the hypotheses in symbols. (2 points)

b) Below is a list of assumptions and conditions. Which of these must be present in order to run the 1-Proportion z-Test. (Write the letter(s) corresponding to the required assumptions and conditions.) (2 points)

A. 10% Condition: I will assume that 500 adults represents less than 10% of the population.

B. Success/Failure Condition: Both ? ∙ ? = 500 ∙ 0.55 = 275 and ? ∙ ? = 500 ∙ 0.45 = 225 are greater than 10.

C. Randomization Condition: The staff conducts the survey randomly.

D. The Nearly Normal Condition: The sample size is 500 so even though we don’t have a histogram of the data it’s reasonable to assume that this condition has been met.

E. Independence Assumption: I will assume that since the survey was done randomly the response of one adult does not affect the response of another adult.

c) Use your calculator to perform a 1-Proportion z-Test and report the test statistic and p-value. Do not make these calculations by hand. Instead, use the 1-PropZTest command in your graphing calculator found under STAT – TESTS. Write out what you entered in your calculator. (3 points)

d) Write a full conclusion for this test in the context of the problem (See #2b for format). (2 points)

Answer 1

(a) = proportion of adults in the school system who approve of the superintendent's performance.
We are to test, (null hypothesis) versus (alternative hypothesis).

(b) All the options A, B, C, D and E are correct.

(c) The commands that we will give under 1-PropZTest (found under STAT - TESTS) are:
.
.
.
.
We then click on "Calculate".
We get the test statistic (z) = 0.8989331 and p-value (p) = 0.1843441.

(d) Since the p-value is greater than the 0.05 significance level, we fail to reject the null hypothesis and conclude that there is not enough evidence to suggest that the approval rating of the superintendent is greater than 55%.