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4 years ago

Can someone please help me with problems 3, 4, and 5?

Mathematics

1 answer:

Romashka [77]4 years ago

6 0

3. i'm not entirely sure about this one, because i feel like i'm making it more complicated than it is, but if there are 100 students and 80% of them would know the answer--80% of 100 is 80. therefore, 80 students would pass and 20 would fail.

4. if you're sending someone an e-mail... the only way to check an email is through the web, so they'll get a 100% "yes i use the web" response because they only sent the survey through e-mail

5. very few, if not absolutely none of the students who received As would ask for a grade re-evaluation. most people are satisfied with As and they'd be scared that, if re-evaluated, their grade would drop. they likely wouldn't respond to the dean in the affirmative.

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3 years ago

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2. A marketing firm is trying to estimate the proportion of potential car buyers that would consider

Maurinko [17]

Answer:

a. The number of people that should be in the pilot study are 600 people

b. The point estimate is 0.62 $\overline 6$

c. At 95% confidence level the true population proportion of potential car buyers of hybrid vehicle is between the confidence interval (0.588, 0.6654)

d. Two ways to reduce the margin of error are;

1) Reduce the confidence interval

2) Use a larger sample size

Step-by-step explanation:

a. The given parameters for the estimation of sample size is given as follows;

The margin of error for the confidence interval, E = 4% = 0.04

The confidence level = 95%

The sample size formula for a proportion as obtained from an online source is given as follows;

$n = \dfrac{Z^2 \times P \times (1 - P)}{E^2}$

Where, P is the estimated proportions of the desired statistic, therefore, we have for a new study, P = 0.5;

Z = The level of confidence at 95% = 1.96

n + The sample size

Therefore, we have;

$n = \dfrac{1.96^2 \times 0.5 \times (1 - 0.5)}{0.04^2} = 600.25$

Therefore, the number of people that should be in the pilot study in order to meet this goal at 95% confidence level is n = 600 people

b. The point estimate for the population proportion is the sample proportion given as follows;

$\hat p = \dfrac{x}{n}$

Where;

x = The number of the statistic in the sample

n = The sample size

From the question, we have;

The number of potential car buyers, n = 600

The number of respondent in the sample that indicated that they would consider purchasing a hybrid, x = 376

Therefore, the point estimate, for the proportion of potential car buyers that would consider buying a hybrid vehicle, $\hat p$ = 376/600 = 0.62 $\overline 6$