Inference for Means Practice

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 t distribution is more accurate than a z distribution.

• Clarity ratings for the entire batch are normally distributed.

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

• The standard deviation for the population is unknown.

Tags

CCSS.HSS.IC.B.4

Thirty randomly selected students took the calculus final. If the sample mean was 92 and the standard deviation was 9.4, construct a 99 percent confidence interval for the mean score of all students from which the sample was gathered.

• 89.08 < μ < 94.92

• 87.27 < μ < 96.73

• 87.29 < μ < 96.71

• 87.77 < μ < 96.23

Tags

CCSS.HSS.IC.B.4

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

• Sample mean
• t*
• Sample standard deviation
• Sample size

Which of the following changes to a study would result in a narrower confidence interval?

• increasing the confidence level, increasing the sample size
• decreasing the confidence level, decreasing the sample size
• increasing the confidence level, decreasing the sample size
• decreasing the confidence level, increasing the sample size.

Tags

CCSS.HSS.IC.B.4

A research group wishes to estimate the mean amount of time (in hours) that members of a fitness center spend exercising each week. They want to estimate the mean within a margin of error of 0.5 hours with a 95% level of confidence. Previous data suggests that the standard deviation of the population is 2.2. Which of the following is the smallest sample size they could use?

• 60

• 75

• 90

• 180

The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were: 13.2,  5.1,  7.5,  8.0,  12.7,  7.6,  13.8,  14.5,  7.7, and 10.5.  Determine a 95 percent confidence interval for the mean time for all players.

• 12.30 < μ < 7.82
•  7.82 < μ < 12.30
• 7.72 < μ < 12.40
• 12.40 < μ < 7.72

Tags

CCSS.HSS.IC.B.4

How do you calculate degrees of freedom?

• n

• n+1

• 1-n

• n-1

A 90% confidence interval for the average salary of all CEOs in the electronics industry was constructed using the results of a random survey of 45 CEOs. The interval was $139,048 to $154,144. Give the interpretation of the interval.

• 90% of the sampled CEOs have salaries that fell in the interval $139,048 to $154,144

• We are 90% confident that the mean salary of all CEOs in the electronics industry falls in the interval $139,048 to $154,144.

• There is a 90% chance that CEOs in the electronics industry have salaries that fall between $139,048 to $154,144

• We are 90% confident that the mean salary of the sampled CEOs falls in the interval $139,048 to $154,144.

Tags

CCSS.HSS.IC.B.4

You sample 12 bugs and find the sample mean is 2.40 cm. You are told the sample standard deviation is 0.2 cm. Find a 95% Confidence Interval.

Identify the Variables

• 12

• n

• 2.40

• s

• 0.2

• t*

• 2.201

• x-bar

Tags

CCSS.HSS.IC.B.4

Find the t* value for a 99% confidence interval with a sample of 23 people.

• 2.807

• 2.508

• 2.064
• 2.819