Past studies have indicated that the percentage of smokers is estimated to be ab
ID: 3055123 • Letter: P
Question
Past studies have indicated that the percentage of smokers is estimated to be about 32%. Given the new smoking cessation programs that have been implemented, you now believe that the percentage of smokers has reduced. You randomly surveyed 2352 people and found that 713 smoke. Use a 0.05 significance level to test the claim that the percentage of smokers has reduced a) Identify the null and alternative hypotheses? Ho: Select an answer+ Hi : | Select an answer # b) What type of hypothesis test should you conduct (left-, right-, or two-tailed)? left-tailed right-tailed two-tailed c) Identify the appropriate significance level. d) Calculate your test statistic. Write the result below, and be sure to round your final answer to two decimal places. e) Calculate your p-value. Write the result below, and be sure to round your final answer to four decimal places. f) Do you reject the null hypothesis? We reject the null hypothesis, since the p-value is less than the significance level. We reject the null hypothesis, since the p-value is not less than the significance level. We fail to reject the null hypothesis, since the p-value is less than the significance level. We fail to reject the null hypothesis, since the p-value is not less than the significance level. g) Select the statement below that best represents the conclusion that can be made. There is sufficient evidence to warrant rejection of the claim that the percentage of smokers is less than 32%. There is not sufficient evidence to warrant rejection of the claim that the percentage of smokers is less than 32%. The sample data support the claim that the percentage of smokers is less than 32%. There is not sufficient sample evidence to support the claim that the percentage of smokers is less than 32%Explanation / Answer
Null Hypothesis: p = 0.32 (stands for p>= 0.32)
Alternate hypothesis: p<0.32
From the above hypothesis, it can be seen that it is a left tailed test.
In this case,
Sample Size n = 2352
Hypothesized value of populaiton propotion P = 0.32
Sample propotion p = 713/2352 = 0.303146
Standard Deviation SD = sqrt[P*(1-P)/n] = 0.010
z score test statistics = (p-P)/SD = -1.75
Since we have a left-tailed test, the p-value is the probability that the z-score is less than -1.75.
Using the Normal Distribution Calculator to find P(z < -1.75) = 0.0401
Thus, the P-value is 0.0401
Thus, it can be seen that P-value (0.0401) is less than the significance level (0.05).
We reject the null hypothesis since p-value is less than the significance level
There is not sufficient evidence to warrent the rejection of the claim that the percentage of smokers is less than 32%