Question

We are testing people to see if the rate of use of seat belts has changed from a previous of 80%. Suppose that in our random sample of 75 people, we see that 66 have the seat belt fastened. The researcher is intersted in a=0.1 level test.

Ste 1. State the null and alternative hypothese.

Step 2. Write down the appropriate test statistiv (formula) and the rejection region of your test (report z critical(s)

Step 3. Compute the value of the test statistic (z observed)

Step 4. State your conclusion (in one sentence, state whether or not the test rejects the null hypothesis and in another sentenve apply the results to the problem).

Step 5. Compute the p- value for this test. State your decision.

Answer 1

Step 1 :

Step 2 :

Test statistic = Z = ( p_o - p ) / ( p_o * ( 1 - p_o) / n )^0.5

Critical value = Z_alpha/2

where , n =75

p = x / n = 66 / 75 = 0.88

Alpha = 0.1

Step 3 :

Test statistic = Z = ( 0.80 - 0.88 ) / ( 0.80 * ( 1- 0.80 ) / 75 )^0.5

Z = - 1.732

Critical value = Z _0.1/2 = Z_0.05 = 1.645

Step 4:

Since,| Z observed | = |-1.732| = 1.732 > 1.645 ( Critical value ) , we reject Ho and conclude that rate of use of seat belts has changed from previous of 80%

Step 5:

Since we have a two-tailed test, the P-value is the probability that the z-score is less than -1.732 or greater than 1.732

We use the Normal Distribution Calculator to find P(z < -1.75) = 0.0416, and P(z > 1.75) = 0.0416 Thus, the P-value = 0.0416 + 0.0416 = 0.0832

P-value = 0.0832 < 0.1 ( Alpha ) , we reject Ho and conclude that rate of use of seat belts has changed from previous of 80%