A random sample is selected from a normal population with a mean of A random sam
ID: 3369527 • Letter: A
Question
A random sample is selected from a normal population with a mean of A random sample is selected fromn a normal population with a mean of ?-30 and a standard deviation of ?-8. After a treatment is administered to the individuals in the sample, the sample mean is found to be M 33. If the sample consists of n 16 scores, is the sample mean sufficient to conclude that the treatment has a significant effect? Use a two-tailed test with a .05. If needed, use the Distributions tool to help you find the critical value. with n = 16, z = You conclude that the treatment has a significant effect. Standard Normal Distribution Mean-0.0 Standard Deviation-1.0 50005000 2.0 1.0 0.0 0.0000 1.0 2.0 ?? If the sample consists of n -64 scores, is the sample mean sufficient to conclude that the treatment has a significant effect? Use a two-tailed test with ?-.os. With n-64, Z_ conclude that the treatment has a significant effect. You _____ Comparing your answers for the two cases you just tested, explain how the size of the sample infiuences the outcome of a hypothesis test A larger sample the likelihood of rejecting the null hypothesis.Explanation / Answer
Solution:-
a)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: u < 30
Alternative hypothesis: u > 30
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.
Analyze sample data. Using sample data, we compute the standard error (SE), and the z statistic test statistic (z).
SE = s / sqrt(n)
S.E = 2.0
z = (x - u) / SE
z = 1.50
where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.
The observed sample mean produced a z statistic test statistic of 1.50.
Thus the P-value in this analysis is 0.067.
Interpret results. Since the P-value (0.067) is greater than the significance level (0.05), we cannot reject the null hypothesis.
We can not conclude that the treatment has a significant effect.
b)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: u < 30
Alternative hypothesis: u > 30
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.
Analyze sample data. Using sample data, we compute the standard error (SE), and the z statistic test statistic (z).
SE = s / sqrt(n)
S.E = 2.0
z = (x - u) / SE
z = 3.0
where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.
The observed sample mean produced a z statistic test statistic of 3.0
Thus the P-value in this analysis is 0.001.
Interpret results. Since the P-value (0.001) is less than the significance level (0.05), we have to reject the null hypothesis.
We can conclude that the treatment has a significant effect.
A larger sample increases the likelihood of rejecting the null hypothesis.
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