Question

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1. (This should look familiar. We had it as an example in module 10.). A business owner wanted to know if the advertisements he has been paying for make people more likely to buy his product. He had a company collect data on people in his target demographic. Use a .03 significance level and the P-value method to test the claim that seeing the ad increases the likelihood of buying the product. [4 pts] Saw the ad Did not see the ad Bought product 140 20 Did not buy product 40 50 Hint: let p = the proportion of people who saw the ad and bought the product and P2 = the proportion of people who did not see the ad and bought the product.

Answer 1

1.

- For sample 1, we have that the sample size is N1 180, the number of favorable cases is X1 = 140, so then the sample proportion is X1 = 0.7778 Pi = X 140 180 = N1 For sample 2, we have that the sample size is N2 = 70, the number of favorable cases is X2 = 20, so then the sample proportion is P2 N = 2 = 0.2857 = The value of the pooled proportion is X1 + X2 140+20 computed as p = 0.64 Ni+N2 180+70 Also, the given significance level is a = 0.03. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: P1 = P2 Ha: 21 > P2 This corresponds to a right-tailed test, for which a z-test for two population proportions needs to be conducted.

(2) Rejection Region Based on the information provided, the significance level is a = 0.03, and the critical value for a right-tailed test is ze = 1.88. The rejection region for this right-tailed test is R = {z: z > 1.88} (3) Test Statistics The Z-statistic is computed as follows: z = 1 – 2 V7(1-7)(1/nı + 1/n2) 0.7778 – 0.2857 10.64. (1 – 0.64)(1/180 + 1/70) 7.278

(4) Decision about the null hypothesis Since it is observed that z = 7.278 > Ze = 1.88, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p= 0, and since p= 0 < 0.03, it is concluded that th null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population proportion p1 is greater than p2, at the 0.03 significance level. Confidence Interval The 97% confidence interval for pı - p2 is: 0.357 < P1 – P2 < 0.627.
Thus there is sufficient evidence to conclude that seeing ad increases the likelihood of buying the product.

2.

We need to construct the 95% confidence interval for the difference between the population means M1 – M2, for the case that the population standard deviations are not known. The following information has been provided about each of the samples: Sample Mean 1 (X) = 10.73 Sample Standard Deviation 1 (81) = 4.25 Sample Size 1 (N1) = 150 Sample Mean 2 (X2) = 13.13 Sample Standard Deviation 2 (S2) = 4.02 Sample Size 2 (N2) = 85 Based on the information provided, we assume that the population variances are equal, so then the number of degrees of freedom are df = n1 + n2 – 2 = 150 + 85 - 2= 233

- ti = The critical value for a = 0.05 and df = 233 degrees of freedom is tc .-a/2;n-1 1.97. The corresponding confidence interval is computed as shown below: Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows: Sp - (n1 – 1)si + (12 – 1)sz ni + n2 2 = (150 – 1) x 4.252 + (85 – 1) x 4.0 150 + 85 – 2 = 4.169 Since we assume that the population variances are equal, the standard error is computed as follows: 1 1 se = + SpV ni n2 1 1 = 4.169 x + 150 85 = 0.566

Now, we finally compute the confidence interval: CI = (X1 - X2 – to x se, X1 - X2 + tc * se = (10.73 – 13.13 – 1.97 x 0.566, 10.73 – 13.13 + 1.97 x 0.566) = (-3.515, -1.285) Therefore, based on the data provided, the 95% confidence interval for the difference between the population means pli – M2 is – 3.515 < Mi – M2 < -1.285, which indicates that we are 95% confident that the true difference between population means is contained by the interval (-3.515, -1.285).

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