Question

significance level a = 0.05.

4. (20 points) A random sample of 1400 Internet users was selected from the records of a large Internet provider and asked whether they would use the Internet or the library to obtain information about health issues. Of these, 872 said they would use the Internet. 1) (10 points) Is there strong evidence that more than half of Internet users would use the Internet to obtain information about health issues? State the Ho and Hq and carry out an appropriate test statistic. Give the P-value and state your conclusion using significance level a = 0.05. 2) (5 points) Construct a 99% confidence interval for the proportion of the Internet users who would use the Internet to obtain information about health issues.

Answer 1

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Let p denotes the population proportion. Hypotheses : To test null hypothesis Ho : p=0.5 against alternative hypothesis Hi :p > 0.5 Let o denotes the sample proportion and n denotes the sample size. Here, .. 872 0.6229, n=1400, standard error = 0.5*(1 -0.5) 1400 0.013363 1400 The test statistic can be written as: ; - po) Po*(1-P) n which under H, follows a standard normal distribution. Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if ZSTAT > 20.05 = 1.644854 0.4- 0.3 0.2 0.1- 0.0 - 3 2 Z Now, Value of test statistic: The value of the test statistic 2STAT = 9.193787 9.194 (0.6229 - 0.5) 0.5*(1-0.5) 1400 The critical value Zcritical = 20.05 = 1.644854 1.645 P-value = P(Z > ZSTAT) = P(Z > 9.193787) =1- P(Z < 9.193787)=1 - 1.000000 = 0.000000 0.0000 Conclusion : Since p-value < 0.05 and ZSTAT > Zcritical = 1.644854, so we reject H, at 0.05 level of significance Hence, we can conclude that the population proportion is significantly greater than 0.5

Let p denotes the population proportion. Let p denotes the sample proportion and n denotes the sample size. Here, 872 PE 0.6229, n= 1400, standard error = 0.623 * (1 – 0.623) 1400 = 0.012953 0.0130 1400 At 99 percent confidence level, Margin of error is p* (1 - ME * 21-0.99 2 0.623 * (1 – 0.623) 1400 * 20.005 = 0.623 *(1 – 0.623) 1400 * 2.575829 = 0.033366 0.033 n a 99 percent confidence interval for the population proportion p is Ô *(1-P) = p = * Z 1-0.99 n = 0.6229 + 0.623 * (1 – 0.623) 1400 * 20.005 = 0.6229 + 0.623 *(1 – 0.623) * 2.575829 1400 = 0.6229 + 0.033366 = (0.589491, 0.656223) ~ (0.589, 0.656)