Question

You are conducting a study to see if the probability of catching the flu this year is significantly more than 0.16 you use a significance level of 0.005 H0: p=0.16 H1:p>0.16 ​ you obtain a sample of size equals 188 in which there are 38 successes what is the P value for this sample? (report answer accurate to four decimal places.) ​ the P value is less than (or equal to) a greater than a ​ this p-value leads to the decision to: reject the null accept the null fail to reject the null ​ as such the final conclusion is that there is sufficient evidence that warrant rejection of the claim that the probability of catching the flu this year is more than 0.16 there is not sufficient evidence to Warrant rejection of the claim that the probability of catching the flu this year is more than 0.16 the sample data support the claim that the probability of catching the flu this year is more than 0.16 ​ there is not sufficient sample evidence to support the claim that the probability of getting the flu this year is more than 0.16​

Answer 1

The p-value for this sample is 0.0165, which is less than the significance level. Therefore, we reject the null hypothesis and conclude that there is sufficient evidence to warrant rejection of the claim that the probability of catching the flu this year is more than 0.16.

Therefore, the correct answer is: option "there is sufficient evidence that warrant rejection of the claim that the probability of catching the flu this year is more than 0.16".

Step-by-step explanation:

The p-value is defined as the probability of obtaining a test statistic as extreme or more extreme than the observed test statistic, assuming the null hypothesis is true.

In this case, the null hypothesis is that the probability of catching the flu this year is 0.16, and the alternative hypothesis is that the probability is greater than 0.16.

To calculate the p-value, we can use the binomial distribution. The test statistic is the z-score, which is calculated as (sample proportion - hypothesized proportion)/sqrt((hypothesized proportion * (1 - hypothesized proportion))/sample size).

In this case, the sample proportion is 38/188 = 0.2021, the hypothesized proportion is 0.16, and the sample size is 188. Plugging these values into the formula, we get a z-score of approximately 2.1316.

Using a standard normal distribution table or a calculator, we can find that the probability of obtaining a z-score of 2.1316 or more extreme is approximately 0.0165. This is the p-value.

Since the p-value is less than the significance level of 0.005, we reject the null hypothesis. The final conclusion is that there is sufficient evidence to warrant rejection of the claim that the probability of catching the flu this year is more than 0.16.