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One-sided test Suppose you want to perform a test of H0: μ=5 versus Ha: μ<5

at the α=0.05 significance level. A random sample of size n=20 from the population of interest yields x¯=4.7 and sx=0.74 . Assume that the conditions for carrying out the test are met.

a. Explain why the sample result gives some evidence for the alternative hypothesis.

b. Calculate the standardized test statistic and P-value.

Short Answer

Expert verified

Part a) The sample mean 4.7is less than 5

Part b)

t=-1.8130.25<P<0.05OrP=0.04283

Step by step solution

01

Part a) Step 1: Given information

H0:μ=5H1:μ<5α=0.05n=20x¯=4.7s=0.74

02

Part a) Step 2: Calculation

The sample mean of 4.7is less than 5,which agrees with the alternative hypothesis that the mean is less than 5,and thus the sample result provides some evidence for the alternative hypothesis.

03

Part b) Step 1: Given information

H0:μ=5H1:μ<5α=0.05n=20x¯=4.7s=0.74

04

Part b) Step 2: Calculation

We know,

t=x¯-μ0sln

The test statistic is

t=x¯-μ0sln=4.7-50.74/20=-1.813

If the null hypothesis is true, the P-value is the probability of getting the test statistic's value or a value that is more extreme.

df=n-1=20-1=19.0.25<P<0.05

Command for Ti83/84-calculator: (-1E99,-1.813,19)which will return a P-value of 0.04283. It could replace -1E99 by any other very small negative number.

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Most popular questions from this chapter

Don't argue Refer to Exercise 2. Yvonne finds that 96 of the 150 students (64%) say they rarely or never argue with friends. A significance test yields a P-value of0.0291 Interpret the P-value.

A company that manufactures classroom chairs for high school students

claims that the mean breaking strength of the chairs is 300 pounds. One of the chairs collapsed beneath a 220-pound student last week. You suspect that the manufacturer is exaggerating the breaking strength of the chairs, so you would like to perform a test of H0:μ=300Ha:μ<300where μ is the true mean breaking strength of this company’s classroom chairs.

a. The power of the test to detect that μ=294 based on a random sample of 30

chairs and a significance level of α=0.05 is 0.71. Interpret this value.

b. Find the probability of a Type I error and the probability of a Type II error for the test in part (a).

c. Describe two ways to increase the power of the test in part (a).

Walking to school Refer to Exercise 36.

a. Explain why the sample result gives some evidence for the alternative hypothesis.

b. Calculate the standardized test statistic and P-value.

c. What conclusion would you make?

A random sample of 100 likely voters in a small city produced 59 voters in favor of Candidate A. The observed value of the standardized test statistic for performing a test of H0:p=0.5H0:p=0.5versus Ha:p>0.5Ha:p>0.5

is which of the following?

a)z=0.59-0.50.59(0.41)100

b)z=0.59-0.50.5(0.5)100

c)z=0.5-0.590.59(0.41)100

d)z=0.5-0.590.5(0.5)100

e)z=0.59-0.5100

Which of choices (a) through (d) is not a condition for performing a significance test about a population proportion p?

a. The data should come from a random sample from the population of interest.

b. Both np0and n(1-p0)should be at least 10.

c. If you are sampling without replacement from a finite population, then you should sample less than 10%of the population.

d. The population distribution should be approximately Normal unless the sample size is large.

e. All of the above are conditions for performing a significance test about a population proportion.

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