# Test hypotheses and sample size requirement

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1. A study was conducted to estimate the mean amount spent on birthday gifts for a typical family having two children. A sample of 150 was taken, and the mean amount spent was $225. Assuming a standard deviation equal to $50, find the 95% confidence interval form, the mean for all such families.

2. In testing the hypothesis, HO: m greater or equal to 28.7 and Ha: m> 28.7, using the p-value approach, a p-value of 0.0764 was obtained. If alpha = 9.8, find the sample mean which produced this p-value given that the sample of size n= 40 randomly selected.

3. To test the null hypothesis that the average lifetime for a particular brand of bulb is 750 hours versus the alternative that the average lifetime is different from 750 hours, a sample of 75 bulbs is used. If the standard deviation is 50 hours and alpha is equal to 0.01, what values for x will result in rejection of the null hypothesis.

4. By measuring the amount of time it takes a component of a product to move from one workstation to the next, and engineer has estimate that the standard deviation is 4.5 seconds. a. How many measurements should be made in order to be 95% certain that the maximum error estimation will not exceed 1 second? b. What sample size is required for a maximum error of 2 seconds?

5. Determine the critical region and critical values for z that would be used to test the null hypothesis at the given level of significance, as described in each of the following:

a. HO:m=25 and Ha: m not equal 25, alpha = 0.10

b. Ho: m< or equal 32 and Ha:m> 32, alpha =0.01

c. Ho: m > or equal 13 and Ha:m< 13, alpha = 0.05

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The solution provides step by step method for calculating confidence intervals and proving the hypothesis tests. The formula for the calculation and interpretations of the results are also included in an attached Word file.

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Please see the attached file for details and missing symbols/formula.

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1. A study was conducted to estimate the mean amount spent on birthday gifts for a typical family having two children. A sample of 150 was taken, and the mena amount spent was $225. Assuming a standard deviation equal to $50, find the 95% confidence interval form, the mean for all such families.

Solution. Given . For a 95% confidence level, z=1.96. So, by a formula, we know that the 95% confidence interval form, the mean for all such families is

=

=(217.00, 233.00)

2. In testing the hypothesis, HO: m greater or equal to 28.7 and Ha: m> 28.7, using the p-value approach, a p-value of 0.0764 was obtained. If alpha = 9.8, find the sample mean which produced this p-value given that the sample of size n= 40 randomly selected. ...

###### Education

- BSc , Wuhan Univ. China
- MA, Shandong Univ.

###### Recent Feedback

- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
- "excellent work"
- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
- "Thank you"
- "Thank you very much for your valuable time and assistance!"

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