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The Institute of Management Accountants (IMA) conducted a survey of senior finance professionals to gauge members’ thoughts on global warming and its impact on their companies. The survey found that 65% of senior professionals that global warming is having a significant impact on the environment. Suppose that you select a sample of 100 senior finance professionals.
1. What is the probability that the sample percentage indicating global warming is having a significant impact on the environment will be between 64% and 69%?
2. The probability is 90% that the sample percentage will be contained within what symmetrical limits of the population percentage?
3. The probability is 95% that the sample percentage will be contained within what symmetrical limits of the population percentage?

Respuesta :

Answer:

(1) The probability that the sample percentage indicating global warming is having a significant impact on the environment will be between 64% and 69% is 0.3674.

(2) The two population percentages that will contain the sample percentage with probability 90% are 0.57 and 0.73.

(3) The two population percentages that will contain the sample percentage with probability 95% are 0.55 and 0.75.

Step-by-step explanation:

Let X = number of senior professionals who thought that global warming is having a significant impact on the environment.

The random variable X follows a Binomial distribution with parameters n = 100 and p = 0.65.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of p if the following conditions are satisfied:

  1. np ≥ 10
  2. n(1 - p) ≥ 10

Check the conditions as follows:

 [tex]np= 100\times 0.65=65>10\\n(1-p)=100\times (1-0.65)=35>10[/tex]

Thus, a Normal approximation to binomial can be applied.

So,  [tex]\hat p\sim N(p, \frac{p(1-p)}{n})=N(0.65, 0.002275)[/tex].

(1)

Compute the value of [tex]P(0.64<\hat p<0.69)[/tex] as follows:

[tex]P(0.64<\hat p<0.69)=P(\frac{0.64-0.65}{\sqrt{0.002275}}<\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}<\frac{0.69-0.65}{\sqrt{0.002275}})[/tex]

                              [tex]=P(-0.20<Z<0.80)\\=P(Z<0.80)-P(Z<-0.20)\\=0.78814-0.42074\\=0.3674[/tex]

Thus, the probability that the sample percentage indicating global warming is having a significant impact on the environment will be between 64% and 69% is 0.3674.

(2)

Let [tex]p_{1}[/tex] and [tex]p_{2}[/tex] be the two population percentages that will contain the sample percentage with probability 90%.

That is,

[tex]P(p_{1}<\hat p<p_{2})=0.90[/tex]

Then,

[tex]P(p_{1}<\hat p<p_{2})=0.90[/tex]

[tex]P(\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}<\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}<\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}})=0.90[/tex]

[tex]P(-z<Z<z)=0.90\\P(Z<z)-[1-P(Z<z)]=0.90\\2P(Z<z)-1=0.90\\2P(Z<z)=1.90\\P(Z<z)=0.95[/tex]

The value of z for P (Z < z) = 0.95 is

z = 1.65.

Compute the value of [tex]p_{1}[/tex] and [tex]p_{2}[/tex]  as follows:

[tex]-z=\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}\\-1.65=\frac{p_{1}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{1}=0.65-(1.65\times 0.05)\\p_{1}=0.5675\\p_{1}\approx0.57[/tex]                 [tex]z=\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}}\\1.65=\frac{p_{2}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{2}=0.65+(1.65\times 0.05)\\p_{1}=0.7325\\p_{1}\approx0.73[/tex]

Thus, the two population percentages that will contain the sample percentage with probability 90% are 0.57 and 0.73.

(3)

Let [tex]p_{1}[/tex] and [tex]p_{2}[/tex] be the two population percentages that will contain the sample percentage with probability 95%.

That is,

[tex]P(p_{1}<\hat p<p_{2})=0.95[/tex]

Then,

[tex]P(p_{1}<\hat p<p_{2})=0.95[/tex]

[tex]P(\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}<\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}<\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}})=0.95[/tex]

[tex]P(-z<Z<z)=0.95\\P(Z<z)-[1-P(Z<z)]=0.95\\2P(Z<z)-1=0.95\\2P(Z<z)=1.95\\P(Z<z)=0.975[/tex]

The value of z for P (Z < z) = 0.975 is

z = 1.96.

Compute the value of [tex]p_{1}[/tex] and [tex]p_{2}[/tex]  as follows:

[tex]-z=\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}\\-1.96=\frac{p_{1}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{1}=0.65-(1.96\times 0.05)\\p_{1}=0.552\\p_{1}\approx0.55[/tex]                 [tex]z=\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}}\\1.96=\frac{p_{2}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{2}=0.65+(1.96\times 0.05)\\p_{1}=0.748\\p_{1}\approx0.75[/tex]

Thus, the two population percentages that will contain the sample percentage with probability 95% are 0.55 and 0.75.

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