Problem A: Confidence Interval Estimates for the Mean

Compute Using: Microsoft Excel functions and/or Data Analysis Toolpak/Excel Workbook files CIE Sigma known.xls and CIE Sigma Unknown.xls

What is the Z-value for 95% level of confidence?

The Z value for 95% level of confidence is calculated using the NORMSINV function provided in the spreadsheet. =NORMSINV((1-B7)/2) which returns -1.9600 using =COMPUTE!B11 formulae.

What is the Z-value for 99% level of confidence?

-2.5758

3. If X = 85, s = 8, and n = 64, construct a 95% confidence interval estimate for the population

mean, m. Explain. (CIE Sigma known.xls)

confidence interval is used to describe the amount of uncertainty associated with sample estimate of a population.

The population mean being constant in the sample this implies that this is not a random variable. The probability is 95% computed to give 11.0019 for the lower limit and 10.9941 for the upper limit.

4. When the population standard deviation is unknown, how can the confidence interval beestimated?

The CI of a population can be calculated when the standard deviation is unknown by using the population mean for the numerical characteristic.

This is achieved by using sample standard deviation s, and/or the sample size from the normal distribution. This is the formulae used

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5. What is the critical value from the t-distribution for a 95% level of confidence?

0.0020

Calculated using =B4/SQRT(B6) formulae.

6. If X = 75 S = 24, and n = 36, assuming that the population is normally distributed, construct a 95% confidence interval estimate for the population mean, m. Explain. (CIE Sigma Unknown.xls)

0.0020

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Problem B: Fundamentals of Hypothesis Testing

Compute Using: Textbook and Excel Workbook files, Z-mean.xls and CIE Sigma Known.xls

Your quality control manager wants to determine whether the mean life of a large shipment of light bulbs is equal to 375 hours. The population standard deviation is 100 hours. A random sample of 64 light bulbs indicates a sample mean life of 350 hours.

At the 0.05 level of significance, is there evidence that the mean life is different from 375 hours?

State the null and alternative hypothesis

Null and alternative hypothesis are mutually exclusive for the 64 light bulbs and the shipment for the hours allocated 100 hours.

Null hypothesis (H0)

The population parameter 64 bulbs and the standard deviation of 100 hours is equal to the 375 hours.

Alternative Hypothesis (H1)

The population parameter of 67 light bulbs combined with the 100 hours standard deviation is less than the 375 hours.

State the decision rule

As a researcher I used the decision rule to decide the acceptance of the null hypothesis stated. I collected the random sample of the 64 light bulbs should be equal to the population mean the sample mean will be acceptable at the standard deviation of 100 hours.

Compute the Z-statistic

-1.95996

Using the formulae =NORMSINV ((1-B7)/2)

State the decision

The mean life of a large shipment of light bulbs is less than 375 hours due to the negative z-value.

Compute the p-value and interpret its meaning.

The expected value is 375 hours divided by 64 bulbs which is 5.859375

Construct a 95% confidence interval estimate of the population mean life of the light bulbs.

Based on your findings, what conclusions do you reach?

The implication of the results after conducting the hypothesis testing indicates that mean life of the shipment of the light bulbs is close to 375 hours.

How is the data useful for effective decision-making?

Problem C: Fundamentals of Hypothesis Testing

Compute Using: Textbook and Excel Workbook files Z-mean.xls and CIE Sigma Known.xls

Your manager wants to determine whether the mean amount of paint contained in 1-gallon cans

purchased from a nationally known manufacturer is actually 1 gallon. You know from the manufacturers specifications that the standard deviation of the amount of paint is 0.02 gallons. You select a random sample of 50 cans, and the mean amount of paint per 1-gallon can is 0.995.

Is there evidence that the mean amount is different from 1.0 gallon (use a = 0.01)?

There is no difference as per the result of 1

State the null and alternative hypothesis

Null Hypothesis: I expect the mean amount of the paint per 1-gallon can to be greater than 1

Alternative Hypothesis: The mean amount of the paint per 1-gallon is less than 1.

State the decision rule

In the 50 samples to be tested the amount of paint per 1-gallon has to be >1 or >0.995.

Compute the Z-test statistic

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The Z-Value is -0.0125

State the decision

The amount of paint per 1-gallon is 1.

Compute the p-value and interpret the meaning.

0.002

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Construct a 99% confidence interval estimate of the population mean amount of paint.

0.00392

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Problem D: Fundamentals of Hypothesis Testing

Compute Using: Textbook, Microsoft Excel functions, and the data provided to complete Problem D.

Consider a sample of n = 16 selected from a normal population, X = 56 and S = 12.

How many degrees of freedom are there in the t-test?

The sample size N=16 degree of freedom = N-1 16-1= 15

3. What are the critical values of t if the level of significance, a, is 0.05 and the alternative

hypothesis, H1, is m 50?

Using formulae =T.DIST(56,15,50) the result of t critical value is 1

What is your statistical decision if the alternative hypothesis, H1 is m 50?

The sample normal population selected to be true is equal to the selected sample.

If in a sample of n = 16, selected from a left skewed population, X = 65 and S = 21, would you use the t-test to test the null hypothesis, H0 = m = 60? Discuss.

No, in this case the sample sizes variances for the two sets are known that is 65 and 21. T-test is used in a situation where the sample size variances are not known.

If, in the sample of n = 160 selected from a left skewed population, X = 65 and S = 21, would you use the t-test to test the null hypothesis, H0 = m = 60. Discuss.

No, in this case the sample sizes variances for the two sets are known that is 65 and 21. T-test is used in a situation where the sample size variances are not known.

Problem E: t-test of Hypothesis for the Mean (The T-Mean.xls is not available)

Compute Using: Textbook and Excel Workbook file T-mean.xls

A manufacturer of chocolate candies uses machines to package candies as they move along a filling line. Although the packages are labeled as 8 ounces, the company wants the package to contain a mean of 8.17 ounces so that virtually none of the packages contain less than 8 ounces. A sample of 50 packages is selected periodically, and the packaging process is stopped if there is evidence that the mean amount packaged is different from 8.17 ounces. Suppose that a particular sample of 50 packages, the mean amount dispersed is 8.159 ounces with a sample standard deviation of 0.051 ounces.

Is there evidence that the population mean amount is different from 8.17 ounces? Use an a = 0.05 level of significance.

a =0.05=5%

State the null and alternative hypotheses

Null hypothesis: I expect the average amount packaged to be more than 8.17 ounces.

H1: m > 8.17

If the amount packaged is not less than 8.17 or greater than 8.17, the amount packaged will remain at 8.159 ounces.

Alternative Hypothesis:

I expect the average amount packaged to be less than 8.17 ounces.

b. State the decision rule

c. Compute the test statistic

d. State the decision

2. Determine the p-value and interpret its meaning

3. Based on your findings, what conclusions do you reach? How is this data useful for effective managerial decision-making?

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