Question

1. Students were asked to look at black and white images and name the objects. The average number named correctly in the US is 20.4 with a standard deviation of 3.2. A researcher wanted to Canadians scored differently on the test. He found that the average score among 30 Canadians was 17.5. Conduct all six steps of hypothesis testing (z-score, 95% confidence level).

Answer 1

Let $\mu$ denotes the average score of Canadians.

Hypotheses : To test null hypothesis Ho u =20.4 against alternative hypothesis H :4+ 20.4 Let i denotes the sample mean, o denotes the population standard deviation and n denotes the sample size. Here, ã = 17.5000,02 = 10.2400,0 = 3.20, n = 30 The test statistic can be written as: Vn(ū – 20.4) which under Ho follows a standard normal distribution. Decision rule / Rejection region : We reject Ho at 0.05 level of significance if P-value < 0.05 or if |zobs > 20.025 = 1.959964 04- 0.3- 20.2- 30.1- 4 -2 0 2 4 Now, Value of test statistic: The value of the test statistic is Zobs = 30 (17.5000 – 20.4) 3.20 = -4.963736 -4.96 The critical value == 1.959964 +1.96 P-value = P(Z > Zobs) = 2 * P(Z < -4.963736) = 2 * 0.000000 = 0.000001 ~ 0.0000 Conclusion: Since p-value < 0.05 and Zobs > critical = 1.96, so we reject Ho at 0.05 level of significance Hence, we can conclude that the the population mean is not equal to 20.4.

Conclusion in problem context :

There is sufficient evidence to support the claim that Canadians scored differently on the test.