Evaluate 26÷2+532−3.

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 $\overline 6$

c. The confidence interval for a proportion is given as follows

$CI=\hat{p}\pm z\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

Therefore, we get;

$CI=0.62 \overline 6\pm 1.96\times \sqrt{\dfrac{\hat{0.62 \overline 6}\cdot (1-\hat{0.62 \overline 6})}{600}}$

C.I. ≈ 0.6267 ± 0.0387

The 95% confidence interval for the true population proportion of potential buyers of hybrid vehicle, C.I. = (0.588, 0.6654)

d. The margin of error is given by the following formula;

$MOE_\gamma = z_\gamma \times \sqrt{\dfrac{\sigma ^2}{n} }$

Where;

$MOE_\gamma$ = Margin of error at a given level of confidence

$z_\gamma$ = z-score

σ = The standard deviation

n = The sample size

Therefore, the margin error can be reduced by the following two ways;

1) Reducing the confidence interval and therefore, the z-score

2) Increasing the sample size

6 0

3 years ago

a culture started with 6000 bacteria after 6 hours it grew to 7200 bacteria predict how many bacteria will be present after 17 h

Zigmanuir [339]

First find the rate of growth using the formula of
A=p e^rt
A 7200
P 6000
E constant
R rate of growth?
T time 6 hours
We need to solve for r
R=[log (A/p)÷log (e)]÷t
R=(log(7,200÷6,000)÷log(e))÷6
R=0.03 rate of growth

Now predict how many bacteria will be present after 17 hours using the same formula
A=p e^rt
A ?
P 6000
R 0.03
E constant
T 17 hours
A=6,000×e^(0.03×17)
A=9,991.7 round your answer to get
A=9992

8 0

3 years ago

Read 2 more answers