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Chapter 9: Q.52 (page 564)

(a) Construct and interpret a 95 % confidence interval for the true proportion p of all first-year students at the university who would identify being well-off as an important personal goal.

(b) Explain what the interval in part (a) tells you about whether the national value holds at this university.

Short Answer

Expert verified

a. The confidence interval is(0.59,0.73)

b. The national value holds at this university as the value lies in the interval

Step by step solution

01

Introduction

A confidence interval is the mean of your estimate in addition to and minus the variation in that estimate. This is the scope of values you anticipate that your estimate should fall between if you re-try your test, within a certain degree of confidence. Confidence, in statistics, is one more method for describing probability.

02

Explanation Part (a)

The number of students is n = 200

Number of students in favour of well being x = 132

population proportion 73%=0.73

p-=xn=132200=0.66

Using,

CI=p-±zα/2×p-(1−p-)n

CI=0.73±1.96×0.73(1-0.73)200

CI=(0.59,0.73)

03

Explanation Part (b)

The national value holds at this university as the value 0.73 lies in the interval.

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Most popular questions from this chapter

A researcher claims to have found a drug that causes people to grow taller. The coach of the basketball team at Brandon University has expressed interest but demands evidence. Over 1000 Brandon students volunteer to participate in an experiment to test this new drug. Fifty of the volunteers are randomly selected, their heights are measured, and they are given the drug. Two weeks later, their heights are measured again. The power of the test to detect an average increase in height of one inch could be increased by

(a) using only volunteers from the basketball team in the experiment.

(b) using A=0.01 instead of A=0.05.

(c) using A=0.05 instead of A=0.01.

(d) giving the drug to 25 randomly selected students instead of 50.

(e) using a two-sided test instead of a one-sided test.

Study more! The significance test in Exercise 76yields a P-value of 0.0622.

(a) Describe a Type I and a Type II error in this setting. Which type of error could you have made in Exercise 76? Why?

(b) Which of the following changes would give the test a higher power to detect μ=120minutes: using α=0.01or α=0.10? Explain.

Tests and CIs The P-value for a one-sided test of the null hypothesisH0:μ=15is 0.03.

(a) Does the 99%confidence interval for μinclude 15? Why or why not?

(b) Does the 95%confidence interval for μinclude 15? Why or why not?

Two-sided test The one-sample t statistic from a sample of n=25observations for the two-sided test of H0:μ=64;Ha:μ≠64has the value t=-1.12.

(a) Find the P-value for this test using (i) Table Band (ii) your calculator. What conclusion would you draw at the 5%significance level? At the 1%significance level?

(b) Redo part (a) using an alternative hypothesis of Ha:μ<64.

Your company markets a computerized device for detecting high blood pressure. The device measures an individual’s blood pressure once per hour at a randomly selected time throughout a 12-hour period. Then it calculates the mean systolic (top number) pressure for the sample of measurements. Based on the sample results, the device determines whether there is significant evidence that the individual’s actual mean systolic pressure is greater than 130. If so, it recommends that the person seek medical attention.

(a) State appropriate null and alternative hypotheses in this setting. Be sure to define your parameter.

(b) Describe a Type I and a Type II error, and explain the consequences of each.

(c) The blood pressure device can be adjusted to decrease one error probability at the cost of an increase in the other error probability. Which error probability would you choose to make smaller, and why?
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