Question

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?

Answer 1

The questions concern calculating the probability of a sample proportion falling within a given range and constructing confidence intervals around the population proportion. These questions are related to statistical concepts such as the normal approximation to the binomial distribution and the use of z-scores for interval estimation, which are typically covered in college-level statistics courses.

Step-by-step explanation:

The student has asked about determining the probability that a sample percentage will fall within certain ranges, given a known proportion from a survey conducted by the Institute of Management Accountants (IMA) regarding the impact of global warming on their companies. Specifically, the student is looking to estimate probabilities related to the sample proportion and construct confidence intervals around a population percentage. These are statistical concepts typically covered in college-level courses in probability and statistics, specifically in chapters related to sampling distributions and confidence interval estimation.

To answer the first question, we would need to use the normal approximation to the binomial distribution, since the sample size is large (n=100). The sample proportion p = 0.65, and we can calculate the standard error for the sampling distribution of the sample proportion. However, since the full calculations are not provided here, a specific numerical answer cannot be given.

For the second and third questions, constructing confidence intervals at 90% and 95% requires using the standard error and the appropriate z-scores that correspond to these confidence levels. Again, the specific limits are not calculated here, but the process involves multiplying the standard error by the z-score and adding and subtracting this product from the sample percentage.

Answer 2

(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:

`np= 100* 0.65=65>10\\n(1-p)=100* (1-0.65)=35>10`

Thus, a Normal approximation to binomial can be applied.

So,
`\hat p\sim N(p, (p(1-p))/(n))=N(0.65, 0.002275)`.

(1)

Compute the value of
`P(0.64<\hat p<0.69)` as follows:

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

`=P(-0.20<Z<0.80)\\=P(Z<0.80)-P(Z<-0.20)\\=0.78814-0.42074\\=0.3674`

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
`p_(1)` and
`p_(2)` be the two population percentages that will contain the sample percentage with probability 90%.

That is,

`P(p_(1)<\hat p<p_(2))=0.90`

Then,

`P(p_(1)<\hat p<p_(2))=0.90`

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

`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`

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

z = 1.65.

Compute the value of
`p_(1)` and
`p_(2)` as follows:

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

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

(3)

Let
`p_(1)` and
`p_(2)` be the two population percentages that will contain the sample percentage with probability 95%.

That is,

`P(p_(1)<\hat p<p_(2))=0.95`

Then,

`P(p_(1)<\hat p<p_(2))=0.95`

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

`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`

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

z = 1.96.

Compute the value of
`p_(1)` and
`p_(2)` as follows:

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

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