Question

An attorney claims that more than 25% of all lawyers advertise. A random sample of 200 lawyers in a certain city showed that 63 had used some form of advertising. At a 0.05, what would be your conclusion regarding the hypothesis test? O Accept the null hypothesis and the test is insignificant. O Reject the null hypothesis and the test is significant O Cannot be determined.

Answer 1

1) Answer. Date: ----21/04/2019 The test hypothesis is The test statistic is -(22.45-18.50)/sqrt(16.4/15+ 18.2/15) 2.6 Given a-0.01, the critical value is lt(0.005, df nl+n2-2-28)-276 (check student t table) Since t-2.6 is less than 2.76, we do not reject Ho. So we can not conclude that there is a difference in average television watching times between the two groups

2) Answer: __ Step 1: State the hypotheses and identify the claim. Step 2: Find the critical values. Since the test is two-tailed and α = 0.05, the degrees offreedom are the smaller of ni- 1 or n2-1 In this case, n-1-10 1-9 and n2 -1-8-1-7. From Table F, the critical values are +2.365 and -2.365 Step 3: Find the Test Value. Change the means to ounces. (1 lb 16 oz) 71b lloz-7×16+ 11-123 oz 71b 4 oz = 7× 16+4=116oz ( x,-X2)-(μι-μ.) (123-116)-0 02.268 3.086 10 8 ธิใ Step 4: Make the decision. Do not reject the null hypothesis 2.365 2.268 +2.365 Step 5: Summarize the results. There is not enough evidence to support the claim that the mean of the weights of the male infants is different from the mean of the weights of the female infants

3) Amesc The summarized data is shown below Children Teens Sample mean X, = 22.45 X2 18.50 Sample variance s 16.4 s-18.2 Sample size n' =15 n2 =15

Let p, be the children (ages 2-11) spend an average of 21 hours 30 minutes watching television per week. Let u, be the children (ages 12-17) spend an average of 20 hours 40 minutes watching television per week. Step 1: The null and alternative hypotheses are defined as follows: The null hypothesis is: Ho: There is no significant difference between the average televisions watching time for the two groups. The alternative hypothesis is H: There is significant difference between the average televisions watching for the two groups. H: (claim) Step 2: Since the test is two-tailed, since α 0.01 , and since the variances are unequal, the degrees of freedom are the smaller of n-1 or n2-1 In this case the degrees of freedom are 15-1-14. Hence from the table the values are +2.977 and -2.977 Step 3: Under the null hypothesis the test statistic is defined as follows: time critical

1n2 (22.45-18.50)-0 16.4 18.2 15 15 3.95 1.5188 = 2.60

Step 4: Consider the following figure: Rejection region Rejection region 2.977 t 2.60 2.977 Making the decision. Here we observe that l (2.60) t2.977), and then we reject the null hypothesis. Step 5: We conclude that there is no enough evidence to support the claim that is significant difference between the average televisions watching do not there time for the two