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Two independent random samples are taken from two populations. The results of these samples are summarized in the next table.

  1. Form a 90% confidence interval for\({\mu _1} - {\mu _2}\)

Short Answer

Expert verified
  1. The confidence interval is (3.59, 4.21)

Step by step solution

01

Given Information

The sample sizes are 135 and 148

The means are 12.2 and 8.3.

The variances are 2.1 and 3.0.

02

Confidence Interval

A confidence interval is an interval centered about the sample statistic with width equal to twice the margin of error. If many sample are taken from a population with the sample size then the proportion of the constructed confidence intervals that will contain the population parameter is\(1 - \alpha \) .

03

Confidence interval for \({\mu _d}\)

The confidence interval computed as

\(\begin{aligned}{l}CI &= \left( {{{\bar x}_1} - {{\bar x}_2}} \right) \pm {z_{\frac{\alpha }{2}}}\left( {\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} } \right)\\ &= \left( {12.2 - 8.3} \right) \pm {z_{0.05}}\left( {\sqrt {\frac{{2.1}}{{135}} + \frac{3}{{148}}} } \right)\\ &= 3.9 \pm 1.645\left( {.189} \right)\\ = 3.9 \pm .310\\ &= \left( {3.9 - .310,3.9 + .310} \right)\\ &= \left( {3.59,4.21} \right)\end{aligned}\)

Therefore, the confidence interval is (3.59, 4.21)

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Most popular questions from this chapter

Drug content assessment. Refer to Exercise 8.16 (p. 467)and the Analytical Chemistry (Dec. 15, 2009) study in which scientists used high-performance liquid chromatography to determine the amount of drug in a tablet. Recall that 25 tablets were produced at each of two different, independent sites. The researchers want to determine if the two sites produced drug concentrations with different variances. A Minitab printout of the analysis follows. Locate the test statistic and p-value on the printout. Use these values α=.05and to conduct the appropriate test for the researchers.

Test and CI for two Variances: Content vs Site

Method

Null hypothesis α1α2=1

Alternative hypothesis α1α2≠1

F method was used. This method is accurate for normal data only.

Statistics

Site N St Dev Variance 95% CI for St Devs

1 25 3.067 9.406 (2.195,4.267)

2 25 3.339 11.147 (2.607,4.645)

Ratio of standard deviation =0.191

Ratio of variances=0.844

95% Confidence Intervals

Method CI for St Dev Ratio CI Variance Ratio

F (0.610, 1.384) (0.372, 1.915)

Tests

Method DF1 DF2 Test statistic p-value

F 24 24 0.84 0.681

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Descriptive Statistics: Support

Variables

N

Mean

StDev

Variance

Minimum

Maximum

Range

Support

992

67.755

26.871

722.036

0.000

155.000

155.000

Assume that σ12=σ22=σ2. Calculate the pooled estimator σ2 for each of the following cases:

a.s12=120,s22=100,n1=n2=25

b.s12=12,s22=20,n1=20,n2=10

c.s12=.15,s22=.20,n1=6,n2=10

d.s12=3000,s22=2500,n1=16,n2=17

Note that the pooled estimate is a weighted average of the sample variances. To which of the variances does the pooled estimate fall nearer in each of the above cases?

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