Hypothesis Testing One Sample

A manufacturer claims that his tires last at least 40,000

miles. A test on 25 tires reveals that the mean life of a tire

is 39,750 miles, with a standard deviation of 387 miles.

Test the Manufacturer’s claim at α = .01.

• Reject Ho

• Reject Ha

• Fail to reject Ho

• Fail to reject Ha

Which of the following shows a right-tailed test?

• Ha: µ < 15

• H₀: µ < 15

• Ha: µ > 15

• H₀: µ > 15

When do you use a t-value instead of a z-value? 

• When n < 30

• When n ≥ 30

• When all of the planets align

• When Ms. O said to

Josh wants to convince his mother to stop buying single-ply toilet paper. Josh believes that even though Fluffy, a two-ply toilet paper costs more, it will last longer because it is more absorbent. To help substantiate his claim, Josh performed a study. He purchased a random sample of 18 rolls of Fluffy. For each roll, he determined how many squares are needed to completely absorb one-quarter cup of water. Shown above is a dotplot of the data. The mean of the sample is 24.444 squares with a standard deviation of 2.45 squares. Single-ply toilet paper requires 26 squares to absorb one-quarter cup of water. Josh would like to carry out a test to determine if there is convincing evidence that the mean number of squares of Fluffy that are needed to absorb one-quarter cup of water is fewer than 26 squares. What is the appropriate test statistic and P-value of this test?

• z = –2.69, P-value = 0. 0036

• t = –2.69, P-value = 0.0078

• t = –2.69, P-value = 0.0156

• t = 2.69, P-value = 0.9922

• z = 2.69, P-value = 0.9964

If we are testing at a 5% significance level and our p-value is 0.00453, what is the conclusion?

• Accept Ho

• Reject Ho

• We need to know chi-squared

• You cannot do this problem without a calculator

A researcher plans to conduct a test of hypotheses at the α = 0.01 significance level.  She designs her study to have a power of 0.90 at a particular alternative value of the parameter of interest.  The probability that the researcher will commit a Type I error is

• 0.01

• 0.10

• 0.90

• 0.99

One ESP (extra sensory perception) test asks the subject to view the backs of cards and identify whether a circle, square, star , or cross is on the front of each card. If p is the proportion of correct answers, this may be viewed as a hypothesis test with H₀: p = 0.25 and H_a: p > 0.25. The subject is recognized to have ESP when the null hypothesis is rejected. What would a Type II error result in?

• Correctly recognizing someone has ESP

• Mistakenly thinking someone has ESP

• Not recognizing that someone really has ESP

• Correctly realizing that someone doesn't have ESP

• Failing to understand the nature of ESP

• Assuming the true difference (Curls – Ponytail) in the mean amount of tips received is 0, there is a 0.0375 probability that Lexie would observe a difference in sample means as greater or greater than she did by chance alone.

• Assuming that the true mean amount of tips that Lexie would receive while wearing curls is greater than the true mean amount of tips that Lexie would receive while wearing her hair in a ponytail, there is 0.0375 probability that the null hypothesis would be rejected.

•  The P-value indicates that there is a probability of 0.0375 that the null hypothesis is true.

•  There is a 0.0375 probability that Lexie will make more money in tips when she wears her hair in curls opposed to wearing her hair in a ponytail.

H₀: μ = k

H₁: μ > k

• Left-tailed test

• Right-tailed test

A level of significance of 5% means:

• There's a 5% chance we're wrong

• There's a 5% chance we'll be wrong if we fail to reject the null hypothesis

• There's a 5% chance we'll be wrong if we reject the null hypothesis.

• There's a 5% chance you'll get an A on the test.