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Confidence Intervals and Hypothesis Test

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The mean and standard deviation of the sample of 65 customer satisfaction ratings are x-bar = 42.95 and s = 2.6424. If we let "mu" denote the mean of all possible customer satisfaction ratings:

a) Calculate 95 and 99 percent confidence intervals for "mu".
b) Using the 95 percent confidence interval, can we be 95% confident that "mu" is greater than 42? Explain>

Suppose we wish to test H0: p = .8 versus Ha: p > .8 and that a random sample of n = 400 gives a
sample proportion = .86.
a Test H0 versus Ha at the .05 level of significance by using a rejection point. What do you conclude?

b Find the p-value for this test.

c Use the p-value to test H0 versus Ha by setting alpha equal to .10, .05, .01, and .001. What do you
conclude at each value of alpha?

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In this solution, formulas, calculation details and explanations are ...

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In this solution, formulas, calculation details and explanations are provided for two problems. The first problem finds 95 and 99 percent confidence intervals for a mean. The second problem conducts a hypothesis test for a mean.

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Confidence Intervals for mean; t test; hypothesis test for proportion

Confidence intervals/One sample hypothesis tests

Directions: For #1, calculate and interpret the confidence interval. Please show your manual calculation or the software output. Please show your work if you calculated manually on #1. If you used statistical software, please show the output.

1: Confidence Intervals for the mean:

A researcher is studying stress among executives. The researcher is using a questionnaire that measures stress. The questionnaire has been validated through past use. A score above 80 indicates stress at a dangerous level. A random sample of 15 executives revealed the following stress level scores.
94
88
73
90
68
79
87
95
87
92
83
94
82
85
84

What is the 95% confidence level? Do you conclude that the executives have a dangerous level of stress? Why?

For #2, which is a one-sample test, identify the null and alternative hypothesis, and the critical value. Then, calculate or identify the test statistic, and make a decision on the null hypothesis. Explain why you made your decision on the null hypothesis

2: Hypothesis test for the population mean: t test
An electronics manufacturing process has a scheduled mean completion time of minutes. It is claimed that, under new management, the mean completion time, , is less than minutes. To test this claim, a random sample of completion times under new management was taken.
The sample had a mean completion time of minutes and a standard deviation of minutes. Assume that the population of completion times under new management is normally distributed. At the level of significance, can it be concluded that the mean completion time, , under new management is less than the scheduled mean?
Perform a one-tailed test.
Hypothesis Test: Mean vs. Hypothesized Value

70.00 hypothesized value
66.00 mean Q1
10.00 std. dev.
2.77 std. error
13 n
12 df

-1.44 t
.0874 p-value (one-tailed, lower)

59.96 confidence interval 95.% lower
72.04 confidence interval 95.% upper
6.04 half-width
1: State the null & alternative hypothesis:
2: Identify the critical value
3: Identify the test statistic
4: State and explain your decision on H0
For #3, which is a one-sample test, identify the null and alternative hypothesis, and the critical value. Then, calculate or identify the test statistic, and make a decision on the null hypothesis. Explain why you made your decision on the null hypothesis

3: Hypothesis test for a population proportion
The manufacturer of a new antidepressant claims that, among all people with depression who use the drug, the proportion of people who find relief from depression is at least . A random sample of patients who use this new drug is selected, and of them find relief from depression. Based on these data, can we reject the manufacturer's claim at the level of significance?
Perform a one-tailed test.
Hypothesis test for proportion vs hypothesized value

Observed Hypothesized
0.7739 0.8 p (as decimal)
178/230 184/230 p (as fraction)
178. 184. X
230 230 n

0.0264 std. error
-0.99 z
.8387 p-value (one-tailed, upper)

Answers:

1: State the null & alternative hypothesis:
2: Identify the critical value
3: Identify the test statistic
4: State and explain your decision on H0

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