Confidence Intervals for Normal and T-Distributions

A local bank needs information concerning the savings account balances of its customers. A random sample of 13 accounts was checked. The mean balance was $686.75 with a sample standard deviation of $256.20. Find a 95% confidence interval for the true mean.

• ($513.14, $860.36)

• ($525.59, $847.91) 

• ($533.27, $840.23)

• ($531.92, $841.58)

What is the t value associated with 90% confidence and a sample size of 15?

• 341

• 345

• 761

• 753

Which of the following does NOT have an affect on the width of a confidence interval?

• Sample mean

• t*

• Sample standard deviation

• Sample size

A 90% confidence interval for the mean speed driven by cars on East Main St. is given by (21, 45). What was the sample mean of this study?

• 12

• 24

• 33

• None of these.

A study surveyed a random sample of 45 students in the high school and found a mean savings of 1800 dollars with a standard deviation of 150 dollars.

Use the normal distribution/empirical rule to estimate a 95% CONFIDENCE INTERVAL for the mean.

*Round to a WHOLE NUMBER

(​ (a)   ​ ,​ (b)   )

• 1755

• 1845

• 1793

• 18007

• 1650

• 1750

For a class project, a student in a statistics class surveys the monthly rent for 32 unfurnished one-bedroom apartments in Boston and finds the mean monthly rent to be $1400 with a standard deviation of $220. Find the 95% confidence interval for the population mean Round to the nearest hundredth

• [1322.22, 1477.78]

• [1395.69, 1404.31]

• [1211.22, 1588.78]

• [905.03, 1894.97]

Why do we use a t-distribution instead of a z-distribution for means?

• Because the population standard deviation is UNKNOWN and it accounts for a greater variability present in SMALL samples.

• It allows us to decrease the margin of error without increasing our confidence level.

• It allows us to use the population standard deviation.

• Because the population standard deviation is KNOWN and it accounts for a greater variability present in LARGE samples.

A recent survey found that 89% of adults pay their bills on time. The margin of error in this survey was 2%. What is the interval of values in which the truth is likely found?

• 89% - 91%

• 87% - 89%

• 87% - 91%

• 388 - 402

A company wishes to determine the cooking time for one pound of spaghetti. The company's technicians cooked one pound of spaghetti and recorded the time needed for the spaghetti to be ready to eat. Repeating this process 35 times resulted in an approximately normal distribution, with a mean of 9.82 minutes and a standard deviation of 1.4 minutes. In which interval should the middle 95% of cooking times fall?

• (8.42, 11.22)

• (7.02, 12.62)

• (9.35, 10.29)

• (6.82, 11.32)

A quality control specialist at a glass factory must estimate the mean clarity rating for a new batch of glass using a sample of 18 glass sheets from the batch. Past investigations show that clarity ratings are normally distributed. The specialist decides to use a t-distribution rather than a z-distribution because ...

• The sample size is too small for a z-distribution

• Clarity ratings for the entire bacth are normally distributed.  

• The t-distribution will create a narrower interval than z.

• The standard deviation for the batch is unknown.