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Chapter 9: Q T8.13 (page 550)

A milk processor monitors the number of bacteria per milliliter in raw milk received at the factory. A random sample of 10one-milliliter specimens of milk supplied by one producer gives the following data:

Construct and interpret a 90% confidence interval for the population mean μ.

Short Answer

Expert verified

The confidence interval is(4794.4,5105.6)

Step by step solution

01

Step 1:Given information

The data is

02

Step 2:Calculation

The formula to compute the confidence interval is:

x¯-ta2,n-1×sn<μ<x¯+ta2,n-1×sn

Follow the provided steps of Minitab to compute the required confidence interval:

1. Enterthe data set in Minitab sheet.

2. Click on Stat > Basic Statistics > 1-Samplet

3. Select Samplein column.

4. Click on options and enter 90%in confidence level.

5. Click OK.

The obtained output is:

\

Hence, the required confidence interval is(4794.4,5105.6)

Therefore,

The obtained confidence interval shows that there is90% probability that the mean number of bacteria per millimeter in raw milk lies between 4794.4 and 5105.6.

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Most popular questions from this chapter

Making conclusions A student performs a test of H0:μ=12versus Ha:μ≠12

at the α=0.05significance level and gets a P-value of 0.01. The

student writes: “Because the P-value is small, we reject H0. The data prove that Hais true.” Explain what is wrong with this conclusion.

Heavy bread? The mean weight of loaves of bread produced at the bakery where you work is supposed to be 1pound. You are the supervisor of quality control at the bakery, and you are concerned that new employees are producing loaves that are too light. Suppose you weigh an SRS of bread loaves and find that the mean weight is 0.975pound.

a. State appropriate hypotheses for performing a significance test. Be sure to define the parameter of interest.

b. Explain why there is some evidence for the alternative hypothesis.

c. The P-value for the test in part (a) is 0.0806. Interpret the P-value.

d. What conclusion would you make at the α=0.01 significance level?

Awful accidents Slow response times by paramedics, firefighters, and policemen can have serious consequences for accident victims. In the case of life-threatening injuries, victims generally need medical attention within 8minutes of the accident. Several cities have begun to monitor emergency response times. In one such city, emergency personnel took more than 8minutes to arrive on 22%of all calls involving life-threatening injuries last year. The city manager shares this information and encourages these first responders to “do better.” After 6months, the city manager selects an SRS of 400 calls involving life- threatening injuries and examines the response times. She then performs a test at the α=0.05level of H0: p=0.22versus Ha:p<0.22, where pis the true proportion of calls involving life-threatening injuries during this 6-month period for which emergency personnel took more than 8minutes to arrive.

a. Describe a Type I error and a Type II error in this setting.

b. Which type of error is more serious in this case? Justify your answer.

c. Based on your answer to part (b), do you agree with the manager’s choice of α=0.05? Why or why not?

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.

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?
See all solutions

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