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According to the Bureau of Labor Statistics, the average age of American workers is 41.9years. The manager of a large technology company believes that the company’s employees tend to be younger, on average. So she takes a random sample of 12 employees and records their ages.

Here are the data:

27 38 32 24 30 47 42 38 27 43 37 33

a. State appropriate hypotheses for testing the manager’s belief. Be sure to define the parameter of interest.

b. State the conditions for performing a test of the hypotheses in (a), and determine whether each condition is met.

c. The P-value of the test is0.003. Interpret this value. What conclusion would you make?

Short Answer

Expert verified

Part a)H0:μ=41.6,H1:μ<41.6

Part b)All conditions are satisfied.

Part c)There is a 0.3%probability of getting the sample results or more extreme, when the mean age of the company's employees is41.9years.

Step by step solution

01

Part (a) Step 1:Given information

Claim is that mean is less than41.6(younger)

02

Part (a) Step 2:Explaination

The null hypothesis statement is that the population value is equal to the given in the claim:

H0:μ=41.6

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis statement is that the null hypothesis is the claim, then the alternative hypothesis statement is the opposite of the null hypothesis.

H1:μ<41.6

μis the mean age of the employees at the company.

03

Part (b)Step 1:Given information

Conditions for performing a test of the hypotheses

04

Part (b)Step 2;Explaination

The three conditions are: Random, independent (10%condition),

Normal/Large sample.

Random: Satisfied, because the sample is a random sample.

Independent: Satisfied, because the sample of 12employees is less than10%of the population of all employees of 12employees is less than 10%of the population of all employees

Normal/ Large sample: Satisfied, because the pattern in the normal quintile plot is roughlyLinear, which indicates that the distribution is approximately Normal

Since all conditions are satisfied, Therefore we can perform a hypothesis test for the population mean

05

Part (c) Step 1:Given information

P=0.003=0.3%

Claim: mean less than41.9.

06

Part (c)Step 2:Explaination

The claim is either the null hypothesis or the alternative hypothesis. The null hypotheses statement is that the population mean is equal to the value given in the claim. If the null hypothesis is the claim, then the alternative hypothesis statement is the opposite of null hypothesis.

H0:μ=41.9

H1:μ<41.9

The P-value is the probability of getting the value of the test statistic or a value more extreme, once the null hypothesis is true.

There is a 0.3% probability of getting the sample results or more extreme, when the mean age of the company's employees is41.9 years.

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Most popular questions from this chapter

You are thinking of conducting a one-sample ttest about a population mean μusing a 0.05significance level. Which of the following statements is correct?

a. You should not carry out the test if the sample does not have a Normal distribution.

b. You can safely carry out the test if there are no outliers, regardless of the sample size.

c. You can carry out the test if a graph of the data shows no strong skewness, regardless of the sample size.

d. You can carry out the test only if the population standard deviation is known.

e. You can safely carry out the test if your sample size is at least 30 .

Tests and confidence intervals The P-value for a two-sided test of the null hypothesis H0:μ=10is0.06

a. Does the 95% confidence interval for μ include 10? Why or why not?

b. Does the 90% confidence interval for μ include 10? Why or why not?

Fast connection? How long does it take for a chunk of information to travel

from one server to another and back on the Internet? According to the site

internettrafficreport.com, the average response time is 200 milliseconds (about one-fifth of a second). Researchers wonder if this claim is true, so they collect data on response times (in milliseconds) for a random sample of 14 servers in Europe. A graph of the data reveals no strong skewness or outliers.

a. State an appropriate pair of hypotheses for a significance test in this setting. Be sure to define the parameter of interest.

b. Check conditions for performing the test in part (a).

c. The 95% confidence interval for the mean response time is 158.22 to 189.64

milliseconds. Based on this interval, what conclusion would you make for a test of the hypotheses in part (a) at the 5% significance level?

d. Do we have convincing evidence that the mean response time of servers in the United States is different from 200 milliseconds? Justify your answer.

Walking to school A recent report claimed that 13%of students typically walk to school. DeAnna thinks that the proportion is higher than 0.13at her large elementary school. She surveys a random sample of 100students and finds that 17typically walk to school. DeAnna would like to carry out a test at the α=0.05significance level of H0:p=0.13versus Ha:p>0.13, where p= the true proportion of all students at her elementary school who typically walk to school. Check if the conditions for performing the significance test are met.

18%Members of the city council want to know if a majority of city residents supports a 1%increase in the sales tax to fund road repairs. To investigate, they survey a random sample of 300city residents and use the results to test the following hypotheses:

H0:p=0.50

Ha:p>0.50

where pis the proportion of all city residents who support a 1%increase in the sales tax to fund road repairs.

In the sample, p^=158/300=0.527, The resulting P-value is 0.18. What is the correct interpretation of this P-value?

a. Only 18% of the city residents support the tax increase.

b. There is an 18%chance that the majority of residents supports the tax increase.

c. Assuming that 50%of residents support the tax increase, there is an 18%probability that the sample proportion would be 0.527or greater by chance alone.

d. Assuming that more than 50%of residents support the tax increase, there is an 18%probability that the sample proportion would be 0.527or greater by chance alone.

e. Assuming that 50%of residents support the tax increase, there is an 18% chance that the null hypothesis is true by chance alone.

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