T Distribution Means

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.

A biologist has taken a random sample of a specific type of fish from a large lake. A 95 percent confidence interval was calculated to be 6.8 ± 1.2 pounds. Which of the following is true? 

• 95 percent of all the fish in the lake weigh between 5.6 and 8 pounds. 
• In repeated sampling, 95 percent of the sample proportions will fall within 5.6 and 8 pounds.
• In repeated sampling, the population mean fish weight will be captured in the interval 95% of the time. 
• There is a 95% chance that the population mean weight is 6.8 pounds.

A researcher wishes to determine the mean energy consumption of a new light bulb. She takes a random sample of 41 bulbs and determines that the mean consumption is 1.3 watts per hour with a standard deviation of 0.7. When constructing a 90% confidence interval, which would be the most appropriate value of the critical value?  

• 1.683
• 1.684
• 1.685
• 1.686

Which is not known in a t-distribution?

• population mean
• population standard deviation
• sample mean
• sample size

These are number of values that are free to vary after a sample statistic has been computed. 

• degrees of feerdom
• degrees of fredome
• degrees of freedome
• degrees of freedom

given n =14 with 99% confidence find the t-value

• 3.010
• 3.011
• 3.012
• 3.013

given n =13 with 95% confidence find the t-value

• 2.200
• 2.201
• 2.202
• 2.203

given n =12 with 90% confidence find the t-value

• 1.796
• 1.797
• 1.798
• 1.799

given df = 30 with 99% confidence find the t-value

• 2.753
• 2.752
• 2.751
• 2.750

given df = 60 with 90% confidence find the t-value

• 1.672
• 1.671
• 1.670
• 1.669