2/5 Problem 2-19 points total Aresearcher came across an article that suggested
ID: 3240652 • Letter: 2
Question
2/5 Problem 2-19 points total Aresearcher came across an article that suggested it was healthy to have more than one close confidant, someone you can discuss matters that are important to you. This ledthe researcher to wonder how many close friends people in his community had, average, and if this average was greater than 1. a. (3 points) Using this information, what is the nul and alternative hypotheses? Write them using statistical notation. b. He was able to gather a random sample of 20 citizens from his communi ity and asked how many close confidants they had. The sample data appeared to be normally shaped with a meanof2.8 and a standard deviation of 1.04 people. (2 points) With this information, describe the sampling distribution of the sample mean, if the null hypothesis is true. (Be sure to comment on the shape, center (if H istrue) and the spread) ii. (1 point) What methods will you be using to finish this hypothesis test, z-methods or t methods? Briefly explain. (4 points) Calculate the appropriate test statistic for this hypothesis test. If necessary, report the degrees of freedom. Show work. 1 point) What is the p-value?Explanation / Answer
Solution:-
a)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: < 1
Alternative hypothesis: > 1
b)
n = 20, xbar = 2.8, = 1.04
i) The sampling distribution of the mean has sample mean of 2.8 and sample standard deviation of 0.233.
ii) We will use t- test for the method becuase population standard deviation is not given and sample is less than 30.
c)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: < 1
Alternative hypothesis: > 1
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. 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 = 0.233
DF = n - 1 = 20 - 1
D.F = 19
t = (x - ) / SE
t = 7.73
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.
d) Thus the P-value in this analysis is 0.00001
Interpret results. Since the P-value (0.00001) is less than the significance level (0.05), we have to reject the null hypothesis.
From the above test we have sufficient evidence in the favor of the claim that people in his community has more than one close friend on average.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.