Question

A publisher reports that 29 % of their readers own a particular make of car. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 250 found that 26% of the readers owned a particular make of car. Is there sufficient evidence at the 0.01 level to support the executive's claim? Step 1 of 7: State the null and alternative hypotheses. Answer. 1 Point Keypad

Answer 1

Let p denotes the true proportion of readers who own a particular make of car.

To test null hypothesis Ho : p = 0.29 against alternative hypothesis H1 :p +0.29 Let o denotes the sample proportion and n denotes the sample size. Here, 0.29 * (1 – 0.29) P = 0.260, n = 250, po = 0.29, standard error of proportion under null = - P=0.028698 250 The test statistic can be written as: | p -p0) 2 = Po*(1-P) п which under Ho follows a standard normal distribution. We reject Ho at 0.01 level of significance if P-value < 0.01 or if |zobs | > 20.005 Now, The value of the test statistic = -1.045353 The critical value == 2.575829 P – value = P(|=| > Zobs) = 2 * P(Z < -1.045353) = 2 * 0.147930 = 0.295860 – 0.2959 Since p-value > 0.01 and |zobs Zcritical = 2.575829, so We fail to reject Ho at 0.01 level of significance Hence, we can conclude that the population proportion is not significantly different from 0.29

Conclusion : i)