Hypothesis Testing & Confidence Interval for Mean

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Children's scores on a measure of aggression are known to have a normal distribution, with a mean of 50 and a standard deviation of 6. A researcher takes a random sample of 49 children, has them listen to calming music for 4 hours, and then measures the aggression level of this sample of children. The sample has a mean of 51. Using a directional hypothesis with a=0.5, complete the following items:
Answers

State the null hypothesis in statistical terms, using statistical notation.
Average aggression level of children ≤ 50 (µ ≤ 50)
State the alternative hypothesis in statistical terms.
Average aggression level of children > 50 (µ > 50)

Define the critical region by drawing the distribution and labelling the z-critical value.
Critical region is obtained from the standard normal distribution at the significance level 0.05.
Critical value = 1.645
Critical region is z > 1.645

Compute the test statistic.
The test statistic used is , where =51, n = 49,  = 6
Therefore, = 1.166666667

what is your decision regarding the null hypothesis?
Do not reject the null hypothesis, since the test statistic is less than the critical value.
Details
Z Test of Hypothesis for the Mean

Data
Null Hypothesis = 50
Level of Significance 0.05
Population Standard Deviation 6
Sample Size 49
Sample Mean 51

Intermediate Calculations
Standard Error of the Mean 0.857142857
Z Test Statistic 1.166666667

Upper-Tail Test
Upper Critical Value 1.644853627
p-Value 0.121672505
Do not reject the null hypothesis

Compute effect size, if appropriate.
Effect size = Mean difference/S.D.
= (51 - 50)/6
= 0.1667

Write a sentence in APA style regarding the results of this study. Include your test statistic (z) written in journal form, as well as any other pertinent statistics.
Do not reject the null hypothesis, since the test statistic is less than the critical value. The sample does not provide sufficient evidence to conclude that the average aggression level of children is more than 50.

Construct 90% confidence interval for the population mean of children's aggression after listening to music, whether it is appropriate or not.
The confidence interval is given by , where = 51, σ = 6, n = 49, = 1.644853627
That is,
= (49.59, 52.41)
Since the calculation of confidence interval involves a two tailed z value, 90% confidence interval is exactly a 5% one tailed test. Hence the 90% confidence interval is appropriate here.
Details
Confidence Interval Estimate for the Mean

Data
Population Standard Deviation 6
Sample Mean 51
Sample Size 49
Confidence Level 90%

Intermediate Calculations
Standard Error of the Mean 0.857142857
Z Value -1.644853627
Interval Half Width 1.409874537

Confidence Interval
Interval Lower Limit 49.59012546
Interval Upper Limit 52.40987454

Interpret your confidence level, that is, report the results of your estimation in a sentence.
With 90% confidence we can claim that population mean of children's aggression after listening to music is within (49.59, 52.41).
Since the confidence interval includes the population mean 50, do not reject the null hypothesis. That is, the hypothesis testing and confidence interval gives the same result.

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