1. Determine the mean, standard deviation and estimated standard error for the a
ID: 3311920 • Letter: 1
Question
1. Determine the mean, standard deviation and estimated standard error for the average reading score for each group of prisoners at the end of the program Prison A. Mean-106.99 SD-29.512 SE-2.853 Prison B. Mean-104.55 SD-25.552 SB-3169 2. State the Null and Alternative Hypotheses for either Prison A or Prison B (the hypotheses will be the same for each prison...) NULL Hypothesis: .-100 * ALTERNATIVE/Research Hypothesis: 100 3. Report the test value and probability of the Null Hypothesis for each sample. . Prison A: Test valueProbability of the Null Hypothesis Prison B: Test value Probubility of he NalliHypothesis 4. Explain your answer to the following questions . Do you accept or reject the Null Hypothesis for Prison A? Explain your answer Do you accept or reject the Null Hypothesis for Prison B? . Explain your answer 5. what is the 95% Confidence Intervals for your sample means? [write a sentence to explain what cach answer means.] . Prison A. . Prison B. . Does the confidence interval inelude zero? (Write a sentence to explain what cach answer means.] Prison A. Prison BExplanation / Answer
Solution:-
1)
Test for prison A
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 100
Alternative hypothesis: 100
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 z-test.
Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).
SE = s / sqrt(n)
S.E = 2.853
z = (x - ) / SE
z = 2.45
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 z statistic is less than -2.45 or greater than 2.45.
Thus, the P-value = 0.0143
Interpret results. Since the P-value (0.0143) is less than the significance level (0.05), we have to reject the null hypothesis.
Test for prison B
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: < 100
Alternative hypothesis: > 100
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 z-test.
Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).
SE = s / sqrt(n)
S.E = 3.169
z = (x - ) / SE
z = 1.44
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.
The observed sample mean produced a z statistic test statistic of 1.44.
Thus the P-value in this analysis is 0.0749
Interpret results. Since the P-value (0.0749) is greater than the significance level (0.05), we cannot reject the null hypothesis.
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