Question

Pollution and altitude: In a random sample of 338 cars driven at low altitudes, 70 of them exceeded a standard of 10 grams of particulate pollution per gallon of fuel consumed. In an independent random sample of 90 cars driven at high altitudes, 17 of them exceeded the standard. Can you conclude that the proportion of high-altitude vehicles exceeding the standard differs from the proportion of low-altitude vehicles exceeding the standard? Let p1 denote the proportion of low-altitude vehicles exceeding the standard and let p2 denote the proportion of high-altitude vehicles exceeding the standard. Use the α=0.10 level of significance and the critical value method.

1.) State the null and alternate hypotheses.

2.)Find the critical value(s). Round the answer to at least three decimal places. If there is more than one critical value, separate them with commas.

3.)Compute the test statistic. Round the answer to two decimal places.

4.) Determine whether to reject H0

5.) State a conclusion.

Answer 1

Two-Proportion Z test

The following information is provided: (a) Sample 1 - The sample size is N1 = 338, the number of favorable cases is X1 = 70 and the sample proportion is p^1​=X1/N1​=70/338​=0.2071 (b) Sample 2 - The sample size is N2 = 90, the number of favorable cases is X2 = 17 and the sample proportion is p^2​=X2/N2​=17/90​=0.1889 and the significance level is α=0.1 Pooled Proportion The value of the pooled proportion is computed as (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: p1 = p2 Ha: p1 ≠ p2 This corresponds to a Two-tailed test, for which a z-test for two population proportions needs to be conducted. (2a) Critical Value Based on the information provided, the significance level is α=0.1, therefore the critical value for this Two-tailed test is Zc​=1.6449. This can be found by either using excel or the Z distribution table. (2b) Rejection Region The rejection region for this Two-tailed test is |Z|>1.6449 i.e. Z>1.6449 or Z<-1.6449 (3) Test Statistics The z-statistic is computed as follows: (4) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is p =P(|Z|>0.3815)=0.7028 (5) The Decision about the null hypothesis (a) Using traditional method Since it is observed that |Z|=0.3815 < Zc​=1.6449, it is then concluded that the null hypothesis is Not rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0.7028, and since p=0.7028>0.1, it is concluded that the null hypothesis is Not rejected. (6) Conclusion It is concluded that the null hypothesis Ho is Not rejected. Therefore, there is Not enough evidence to claim that the population proportion p1 is different than p2, at the 0.1 significance level.