Question

Tweet tweet: An article reported that 73% of companies have Twitter accounts. An economist thinks the percentage is lower at technology companies. She samples 70 technology companies and finds that 43 of them have Twitter accounts. Can she conclude that less than 73% of technology companies have Twitter accounts? Use the a= 0.01 level of significance and the P-value method. Part: 0 / 5 Part 1 of 5 State the null and alternate hypotheses. HO: H: This hypothesis test is a (Choose one) test.

B ) IF MAKING A DECISION, CHOOSE A SIGNIFICANCE LEVEL

C ) COMPUTE THE TEST STATISTIC

D) COMPUTE THE P VALUE

E ) IS THERE OR IS THERE NOT ENOUGH EVIDENCE TO CONCLUDE MORE THAN 73% TWITTER ACCOUNTS

Answer 1

The following information is provided: The sample size is N = 70, the number of favorable cases is X = 43, and the sample proportion is p = * = * = 0.6143, and the significance level is a = 0.01 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho:p= 0.73 Ha: p < 0.73 This corresponds to a left-tailed test, for which a z-test for one population proportion needs to be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.01, and the critical value for a left-tailed test is ze = -2.33. The rejection region for this left-tailed test is R = {z: z< -2.33} (3) Test Statistics The Z-statistic is computed as follows: P - PO V po(1 – Po)/n 0.6143 -0.73 70.73(1 – 0.73)/70 -2.181 (4) Decision about the null hypothesis Since it is observed that z = -2.181 > Zc = -2.33, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.0146, and since p = 0.0146 > 0.01, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population proportion p is less than po, at the a = 0.01 significance level. Confidence Interval The 99% confidence interval for p is: 0.464 < p < 0.764. Graphically. Z-Test Results: z-stats = -2.181, p-value =0.0146 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 -4.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

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