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Chapter 7: Q40E (page 303)

presented a sample of \({\rm{n = 153}}\)observations on ultimate tensile strength, the previous section gave summary quantities and requested a large-sample confidence interval. Because the sample size is large, no assumptions about the population distribution are required for the validity of the CI.

a. Is any assumption about the tensile-strength distribution required prior to calculating a lower prediction bound for the tensile strength of the next specimen selected using the method described in this section? Explain.

b. Use a statistical software package to investigate the plausibility of a normal population distribution.

c. Calculate a lower prediction bound with a prediction level of \({\rm{95\% }}\)for the ultimate tensile strength of the next specimen selected.

Short Answer

Expert verified

a) Assume normality

b) The population distribution is not normal

c) The lower prediction bound is \(127.7689.\)

Step by step solution

01

To investigate the plausibility of a normal population distribution

(a):

Yes. The assumption that the sample is from the normally distributed population is needed.

Hence Assume normality.

(b):

The normal probability plot suggests that the population distribution is not normal. The indicator of this is that the data does not lay on the line.

Hence the population distribution is not normal

02

To calculate a lower prediction bound

(c):

The mean of the sample has been calculated and it is

\({\rm{\bar x = 135}}{\rm{.39}}\)

and the sample standard deviation

\({\rm{s = 4}}{\rm{.59}}\)

where \({\rm{n = 153}}\). The t value, for \({\rm{\alpha = 0}}{\rm{.05}}\)which was obtained from

\({\rm{100(1 - \alpha ) = 95}}\)is

\({{\rm{t}}_{{\rm{\alpha ,n - 1}}}}{\rm{ = }}{{\rm{t}}_{{\rm{0}}{\rm{.05,152}}}}{\rm{ = 1}}{\rm{.655}}\)

where the t value was computed using a computer software.

The lower prediction bound can be computed using formula

\({\rm{\bar x - }}{{\rm{t}}_{{\rm{\alpha /2,n - 1}}}}{\rm{ \times s}}\sqrt {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{n}}}} \)

Therefore, the lower prediction bound is

\({\rm{135}}{\rm{.39 - 1}}{\rm{.655 \times 4}}{\rm{.59 \times }}\sqrt {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{153}}}}} {\rm{ = 127}}{\rm{.7689}}\)

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Most popular questions from this chapter

Consider the next \({\rm{1000}}\) 95% CIs for m that a statistical consultant will obtain for various clients. Suppose the data sets on which the intervals are based are selected independently of one another. How many of these\({\rm{1000}}\)intervals do you expect to capture the corresponding value of m? What is the probability that between \({\rm{940 and 960}}\)Do these intervals contain the corresponding value of m? (Hint: Let Y = the number among the \({\rm{1000}}\) intervals that contain m. What kind of random variable is Y?)

The negative effects of ambient air pollution on children鈥檚 lung function has been well established, but less research is available about the impact of indoor air pollution. The authors of 鈥淚ndoor Air Pollution and Lung Function Growth Among Children in Four Chinese Cities鈥� investigated the relationship between indoor air-pollution metrics and lung function growth among children ages \({\rm{6--13}}\)years living in four Chinese cities. For each subject in the study, the authors measured an important lung-capacity index known as FEV1, the forced volume (in ml) of air that is exhaled in \({\rm{1}}\) second. Higher FEV1 values are associated with greater lung capacity. Among the children in the study, \({\rm{514}}\) came from households that used coal for cooking or heating or both. Their FEV1 mean was \({\rm{1427}}\)with a standard deviation of \({\rm{325}}\). (A complex statistical procedure was used to show that burning coal had a clear negative effect on mean FEV1 levels.)

a. Calculate and interpret a \({\rm{95\% }}\) (two-sided) confidence interval for true average FEV1 level in the population of all children from which the sample was selected. Does it appear that the parameter of interest has been accurately estimated?

b. Suppose the investigators had made a rough guess of \({\rm{320}}\) for the value of s before collecting data. What sample size would be necessary to obtain an interval width of \({\rm{50}}\)ml for a confidence level of \({\rm{95\% }}\)?

Aphid infestation of fruit trees can be controlled either by spraying with pesticide or by inundation with ladybugs. In a particular area, four different groves of fruit trees are selected for experimentation. The first three groves are sprayed with pesticides 1, 2, and 3, respectively, and the fourth is treated with ladybugs, with the following results on yield:

Treatment ni = Number of Trees (Bushels/Tree) si

1 100 10.5 1.5

2 90 10.0 1.3

3 100 10.1 1.8

4 120 10.7 1.6

Let 碌i= the true average yield (bushels/tree) after receiving the ith treatment. Then

\({\bf{\theta = }}\frac{{\bf{1}}}{{\bf{3}}}{\bf{(}}{{\bf{\mu }}_{\bf{1}}}{\bf{ + }}{{\bf{\mu }}_{\bf{2}}}{\bf{ + }}{{\bf{\mu }}_{\bf{3}}}{\bf{) - }}{{\bf{\mu }}_{\bf{4}}}\)

measures the difference in true average yields between treatment with pesticides and treatment with ladybugs. When n1, n2, n3, and n4 are all large, the estimator \(\widehat {\bf{\theta }}\) obtained by replacing each \({{\bf{\mu }}_{\bf{i}}}\)by Xiis approximately normal. Use this to derive a large-sample 100(1 -伪)% CI for \({\bf{\theta }}\), and compute the 95% interval for the given

data.

Let X1, X2,鈥�, Xn be a random sample from a continuous probability distribution having median \(\widetilde {\bf{\mu }}\) (so that \({\bf{P(}}{{\bf{X}}_{\bf{i}}} \le \widetilde {\bf{\mu }}{\bf{) = P(}}{{\bf{X}}_{\bf{i}}} \le \widetilde {\bf{\mu }}{\bf{) = }}{\bf{.5}}\)).

a. Show that

\({\bf{P(min(}}{{\bf{X}}_{\bf{i}}}{\bf{) < }}\widetilde {\bf{\mu }}{\bf{ < max(}}{{\bf{X}}_{\bf{i}}}{\bf{)) = 1 - }}{\left( {\frac{{\bf{1}}}{{\bf{2}}}} \right)^{{\bf{n - 1}}}}\)

So that \({\bf{(min(}}{{\bf{x}}_{\bf{i}}}{\bf{),max(}}{{\bf{x}}_{\bf{i}}}{\bf{))}}\)is a \({\bf{100(1 - \alpha )\% }}\) confidence interval for \(\widetilde {\bf{\mu }}\) with (\({\bf{\alpha = 1 - }}{\left( {\frac{{\bf{1}}}{{\bf{2}}}} \right)^{{\bf{n - 1}}}}\).Hint :The complement of the event \(\left\{ {{\bf{min(}}{{\bf{X}}_{\bf{i}}}{\bf{) < \mu < max(}}{{\bf{X}}_{\bf{i}}}{\bf{)}}} \right\}\)is \({\bf{\{ max(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \le \widetilde {\bf{\mu }}{\bf{\} }} \cup {\bf{\{ min(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \ge \widetilde {\bf{\mu }}{\bf{\} }}\). But \({\bf{max(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \le \widetilde {\bf{\mu }}\) iff \({\bf{(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \le \widetilde {\bf{\mu }}\)for all i.

b.For each of six normal male infants, the amount of the amino acid alanine (mg/100 mL) was determined while the infants were on an isoleucine-free diet,

resulting in the following data:

2.84 3.54 2.80 1.44 2.94 2.70

Compute a 97% CI for the true median amount of alanine for infants on such a diet(鈥淭he Essential Amino Acid Requirements of Infants,鈥� Amer. J. Of Nutrition, 1964: 322鈥�330).

c.Let x(2)denote the second smallest of the xi鈥檚 and x(n-1) denote the second largest of the xi鈥檚. What is the confidence level of the interval(x(2), x(n-1)) for \(\widetilde \mu \)?

Unexplained respiratory symptoms reported by athletes are often incorrectly considered secondary to exercise-induced asthma. The article 鈥淗igh Prevalence of Exercise-Induced Laryngeal Obstruction in Athletes鈥� (Medicine and Science in Sports and Exercise, 2013: 2030鈥�2035) suggested that many such cases could instead be explained by obstruction of the larynx. In a sample of 88 athletes referred for an asthma workup, 31 were found to have the EILO condition.

a.Calculate and interpret a confidence interval using a 95% confidence level for the true proportion of all athletes found to have the EILO condition under these circumstances.

b.What sample size is required if the desired width of the 95% CI is to be at most .04, irrespective of the sample results?

c.Does the upper limit of the interval in (a) specify a 95% upper confidence bound for the proportion being estimated? Explain.

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