Suppose that, in 1995, the weight of 16-year old males in the United States was normally distributed with a mean of 143 lbs and standard deviation equal to 15 lbs. Suppose that Roger, a public health researcher, believes that the popularity of video and computer gaming has made teenagers less active, and predicts that 16-year old males today weigh more than 16-year old males did in 1995. To test the theory, he chooses a random sample of 50 16-year old males and records their weights. The average weight of the sample is 147 lbs. Roger uses a z ‑ test to test the hypothesis that the average weight of 16-year old males has increased. He calculates the z ‑ statistic as 1.89 standard deviations above the mean. Use this standard normal z-distribution table to calculate the p-value of this statistic. Give your answer as a decimal rounded to four places. Also how to determin if this is a one-sided or two-sided test?
Accepted Solution
A:
Answer:Step-by-step explanation:The null hypothesis is:H0: μ(1995)=μ(2019)The alternative hypothesis is:H1: μ(1995)<μ(2019)Because Roger wants to know if mean weight of 16-old males in 2019 is more than the mean weight of 16-old males in 1995 the test only uses one tail of the z-distribution. It is not a two-sided test because in that case the alternative hypothesis would be: μ(1995)≠μ(2019).To know the p-value, we use the z-statistic, in this case 1.89 and the significance level. Because the problem does not specify it, we will search for the p-value at a 5% significance level and at a 1%.For a z of 1.89 and 5% significance level, the p-value is: 0.9744For a z of 1.89 and 1% significance level, the p-value is: 0.9719