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# Confidence intervals for estimation of parameters

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Question 1
1. A random sample of 100 students attending a concert spent an average of \$142 on their tickets with a standard deviation of \$47.50.

Calculate the 90%, 95%, and 99% confidence intervals for the mean amount of money spent by all students attending the concert. Interpret your response within the context of the situation. Refer to Chapter 17, pp. 427-430 on calculating confidence intervals.
Zikmund, W. G., Babin, B. J., Carr, J. C., & Griffin, M. (2013). Business research
methods. (9th ed.). Mason, OH: South-Western.

Your response should be at least 75 words in length, unless otherwise specified. You are required to use at least your textbook as source material for your response. All sources used, including the textbook, must be referenced; paraphrased and quoted material must have accompanying citations.
Question 2
1. Keri is the owner of a new restaurant in the downtown area of her hometown. To continuously improve service, she would like to know if completed dishes are being delivered to the customer's table within one minute of being completed by the chef. A random sample of 75 completed dishes showed that 60 were delivered within one minute of completion.

Calculate the 90%, 95%, and 99% confidence interval for the true population proportion. Interpret your response within the context of the situation. Refer to Chapter 17, pp. 427-430 on calculating confidence intervals.

Zikmund, W. G., Babin, B. J., Carr, J. C., & Griffin, M. (2013). Business research
methods. (9th ed.). Mason, OH: South-Western.

Your response should be at least 75 words in length, unless otherwise specified. You are required to use at least your textbook as source material for your response. All sources used, including the textbook, must be referenced; paraphrased and quoted material must have accompanying citations.

#### Solution Preview

Question 1

Identify the following values:
n = 100
sample mean = 142
s = 47.5

For 90% confidence interval,
critical value z = 1.645

E = 1.645 (47.5/sqrt(100)) = 0.781

142 - 0.781 = 141.219
142 + 0.781 = 142.781

We are 90% confident that the true mean amount of money spent by all students attending the concert is between \$141.22 and \$142.78

For 95% confidence interval,
critical value z = 1.96

E = 1.96 (47.5/sqrt(100)) = 0.931

142 - 0.931 = 141.069
142 + 0.931 = 142.931

We are 95% ...

#### Solution Summary

A step-by step explanation with computations has been provided in 342 words.

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