Question

A local car dealer claims that 25% of all cars in San Francisco are blue. You take a random sample of 600 cars in San Francisco and find that 141 are blue. Conduct a hypothesis test to see if you can reject the dealer's claim with a significance level of 0.05. Include all ten steps of the hypothesis testing process.

Write your answer on a piece of paper and upload a photo or scan, or else type your answer into a electronic document and upload that document.

In order to get full credit you must include all of the following in your answer:

a. Null and alternative hypotheses stated in mathematical terms.
b. The symbol for the test statistic you used to evaluate your hypothesis test.
c. The value of the test statistic you used to evaluate your hypothesis test.
d. The p-value for your test.
e. The statistical result of your test (did you reject or fail to reject?)
f. A sentence explaining the result of your test in business terms rather than mathematical terms (in other words, how would you explain the meaning of this result to someone who has not studied statistics, like your boss?)

Answer 1

Explanation:

a) We would set up the hypothesis test.

For the null hypothesis,

p = 0.25

For the alternative hypothesis,

p ≠ 0.25

b) The test statistic to be used is the z test. The formula is

z = (P - p)/√pq/n

c) Considering the population proportion, probability of success, p = 0.25

q = probability of failure = 1 - p

q = 1 - 0.25 = 0.75

Considering the sample,

Sample proportion, P = x/n

Where

x = number of success = 141

n = number of samples = 600

P = 141/600 = 0.235

Therefore

z = (0.235 - 0.25)/√(0.25 × 0.75)/600 = - 0.85

d) Recall, population proportion, p = 0.25

The difference between sample proportion and population proportion is 0.25 - 0235. = 0.015

Since the curve is symmetrical and it is a two tailed test, the p for the left tail is 0.25 - 0.015 = 0.235

the p for the right tail is 0.25 + 0.015 = 0.265

These proportions are lower and higher than the null proportion. Thus, they are evidence in favour of the alternative hypothesis. We will look at the area in both tails. Since it is showing in one tail only, we would double the area

From the normal distribution table, the area below the test z score in the left tail 0.198

We would double this area to include the area in the right tail of z = 0.85. Thus

p = 0.198 × 2 = 0.396

e) Since alpha, 0.01 < than the p value, 0.396, then we would fail to reject the null hypothesis.

f) In a sample of 600 cars, we would observe a sample proportion of 0.015 or more away from 0.25 about 39.6% of the time by chance alone. Therefore, the sample result is not statistically significant because it could occur in 39.6% of the time by chance alone.