This case study is about a study on food consumption habits of teenagers. In tot
ID: 3200425 • Letter: T
Question
This case study is about a study on food consumption habits of teenagers. In total, 41 teenagers were randomly selected and asked how many 12-ounces servings of soda he or she drinks each day. Of these 41, 16 were females and 25 were males.1. For the entire random sample of 41 male and female teenagers, the sample mean is 1.892 servings. Assume the population standard deviation is 0.986 servings and that the population is normally distributed. Is there enough evidence to support a claim that the mean amount of servings all teenagers drink is not exactly 1.75 per day? Assume = 0.04. This case study is about a study on food consumption habits of teenagers. In total, 41 teenagers were randomly selected and asked how many 12-ounces servings of soda he or she drinks each day. Of these 41, 16 were females and 25 were males.
1. For the entire random sample of 41 male and female teenagers, the sample mean is 1.892 servings. Assume the population standard deviation is 0.986 servings and that the population is normally distributed. Is there enough evidence to support a claim that the mean amount of servings all teenagers drink is not exactly 1.75 per day? Assume = 0.04. This case study is about a study on food consumption habits of teenagers. In total, 41 teenagers were randomly selected and asked how many 12-ounces servings of soda he or she drinks each day. Of these 41, 16 were females and 25 were males.
Explanation / Answer
Solution:
1)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 1.75
Alternative hypothesis: 1.75
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.04. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n)
S.E = 0.15399
DF = n - 1 = 41 - 1 = 40
t = (x - ) / SE
t = 0.92214
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
Since we have a two-tailed test, the P-value is the probability that the t statistic having 40 degrees of freedom is less than - 0.922 or greater than 0.922
We use the t Distribution Calculator to find P(t < - 0.922) = 0.181, and P(t > 0.922) = 0.181
Thus, the P-value = 0.181 + 0.181 = 0.362
Interpret results. Since the P-value (0.362) is greater than the significance level (0.04), we cannot reject the null hypothesis.
From this we can conclude that we do not have sufficient evidence in the favor of the claim that mean amount of serving is not exactly 1.75 per day.