9.5 Additional information and full hypothesis test examples  (Page 6/51)

Let’s follow a four-step process to answer this statistical question.

1. State the Question : We need to determine if, at a 0.05 significance level, the average conductivity of the selected glass is greater than one. Our hypotheses will be
   1. H ₀ : μ ≤ 1
   2. H _a : μ >1
2. Plan : We are testing a sample mean without a known population standard deviation. Therefore, we need to use a Student's-t distribution. Assume the underlying population is normal.
3. Do the calculations : We will input the sample data into the TI-83 as follows.

   

   

   

   

4. State the Conclusions : Since the p -value* ( p = 0.036) is less than our alpha value, we will reject the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one.

We will follow the four-step process.

1. We need to conduct a hypothesis test on the claimed cancer rate. Our hypotheses will be
   1. H ₀ : p ≤ 0.00034
   2. H _a : p >0.00034

   If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer.

2. We will be testing a sample proportion with x = 172 and n = 420,019. The sample is sufficiently large because we have np = 420,019(0.00034) = 142.8, nq = 420,019(0.99966) = 419,876.2, two independent outcomes, and a fixed probability of success p = 0.00034. Thus we will be able to generalize our results to the population.
3. The associated TI results are
   

   

   

4. Since the p -value = 0.0073 is greater than our alpha value = 0.005, we cannot reject the null. Therefore, we conclude that there is not enough evidence to support the claim of higher brain cancer rates for the cell phone users.

We will follow the four-step plan.

1. We need to test whether the proportion of sexual assaults in Daviess County, KY is significantly different from the national average.
2. Since we are presented with proportions, we will use a one-proportion z -test. The hypotheses for the test will be
   1. H ₀ : p = 0.00078
   2. H _a : p ≠ 0.00078
3. The following screen shots display the summary statistics from the hypothesis test.

   

   

4. Since the p -value, p = 0.00063, is less than the alpha level of 0.01, the sample data indicates that we should reject the null hypothesis. In conclusion, the sample data support the claim that the proportion of sexual assaults in Daviess County, Kentucky is different from the national average proportion.