A sample mean, sample standard deviation, and sample size are given. Use the one
ID: 3226862 • Letter: A
Question
A sample mean, sample standard deviation, and sample size are given. Use the one mean t-test to perform the required hypothesis test about the mean, mu, of the population from which the sample was drawn. Use the critical-value approach. x bar = 20.5, s = 7, n = 11, H_0 =: mu = 18.7, H_a: mu not equal to 18.7, alpha = 0.05 Test statistic: t = 0.85. Critical values t = plusminus 12.228. Do not reject H_0. There is not sufficient evidence to conclude that the mean is different from 18.7. Test statistic: t = 0.85. Critical values: t = plusminus 2.201. Do not reject H_0. There is not sufficient evidence to conclude that the mean is different from 1.87. Test statistic: t = 0.85. Critical values: t = plusminus 2.201. Reject H_0. There is sufficient evidence to conclude that the mean is different from 18.7. Test statistic: t = 0.85. Critical values: t = plusminus 1.96. Reject H_0. There is sufficient evidence to conclude that the mean is different from 187.Explanation / Answer
Solution:- (A)
The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 18.7
Alternative hypothesis: 18.7
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 0.05. 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) = 7 / sqrt(11) = 2.11
DF = n - 1 = 11 - 1 = 10
t = (x - ) / SE = (20.5 - 18.7)/2.11 = 0.85
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.
We use the t Distribution Calculator to find P(t < 0.85)
The P-Value is 0.415.
The result is not significant at p < 0.05.
Critical value : t = + 2.228
Interpret results. Since the P-value (0.415) is greater than the significance level (0.05), we cannot reject the null hypothesis.
There is not sufficient evidence to conclude that the mean is different from 18.7.