Question #175446

1.Draw a normal curve showing the 99% confidence interval

2.Appliace manufacturers are required to post a sticker on their products regarding the electricity economy of each appliance for sale. Explain how this sticker indicates estimation in general.

3.Compare and contrast the z-distribution and the t-distribution.

4.Using t-table, give the confidence coefficients for each of the following:

a. n = 12 with 95% confidence

b. n = 15 with 95% confidence

c. n = 21 with 99% confidence

d. n = 23 with 95% confidence

e. n = 25 with 99% confidence

5.Assuming that the samples come from normal distributions, find the margin of error E given the following:

a. n = 10 and X = 28 with s = 4.0, 90% confidence

b. n = 16 and X = 50 with s = 4.2, 95% confidence

c. n = 20 and X = 68.2 with s = 2.5, 90% confidence

d. n = 23 and X = 80.6 with s = 3.2, 95% confidence

e. n = 25 and X = 92.8 with s = 2.6, 99% confidence

6.Using the information in number 3, find the interval estimates of the population mean.

a.

b.

c.

d.

e.


1
Expert's answer
2021-03-31T13:20:03-0400

1.



2.

Dark green, it represents consumption of less than 25%.

With its intermediate green colour, it symbolizes consumption of less than 30%.

Its pale green colour represents consumption estimated as from 30% to 42%.

Yellow in colour, it offers average consumption between 42% and 55%.

Identified by the colour orange, it represents consumption of between 55% and 75%.

Bright orange in colour, it represents a consumption level of between 75% and 90%

Red in colour, it indicates that it has a consumption level of between 90% and 100%.


3.

The distribution curves are both​ mound-shaped and symmetric. ​ However, the​ t-test statistic curve is flatter than the​ z-statistic curve because the​ t-test is much more sensitive to the sample size. The standard normal or z-distribution assumes that you know the population standard deviation. The t-distribution is based on the sample standard deviation.


4.a) t=1.78t=1.78

b) t=1.75t=1.75

c) t=2.52t=2.52

d) t=1.71t=1.71

e) t=2.49t=2.49


5.

E=zαsnE=z_{\alpha}\frac{s}{\sqrt{n}}

a.

E=1.65410=2.087E=1.65\cdot\frac{4}{\sqrt{10}}=2.087

b.

E=1.964.216=2.058E=1.96\cdot\frac{4.2}{\sqrt{16}}=2.058

c.

E=1.652.520=0.922E=1.65\cdot\frac{2.5}{\sqrt{20}}=0.922

d.

E=1.963.223=1.308E=1.96\cdot\frac{3.2}{\sqrt{23}}=1.308

e.

E=2.582.625=1.3416E=2.58\cdot\frac{2.6}{\sqrt{25}}=1.3416


6.

zα/2<Xμs/n<z1α/2z_{\alpha/2}<\frac{X-\mu}{s/\sqrt{n}}<z_{1-\alpha/2}

(Xzα/2s/n)>μ>(Xz1α/2s/n)(X-z_{\alpha/2}\cdot s/\sqrt{n})>\mu>(X-z_{1-\alpha/2}\cdot s/\sqrt{n})

a.

(28+2.5754/10)>μ>(281.6454/10)(28+2.575\cdot4/\sqrt{10})>\mu>(28-1.645\cdot4/\sqrt{10})

25.92<μ<31.2625.92<\mu<31.26

b.

(501.964.2/16)<μ<(50+1.964.2/16)(50-1.96\cdot4.2/\sqrt{16})<\mu<(50+1.96\cdot4.2/\sqrt{16})

47.94<μ<52.0647.94<\mu<52.06

c.

(68.21.6452.5/20)<μ<(68.2+2.5752.5/20)(68.2-1.645\cdot2.5/\sqrt{20})<\mu<(68.2+2.575\cdot2.5/\sqrt{20})

67.28<μ<69.6467.28<\mu<69.64

d.

(80.61.963.2/23)<μ<(80.6+1.963.2/23)(80.6-1.96\cdot3.2/\sqrt{23})<\mu<(80.6+1.96\cdot3.2/\sqrt{23})

79.29<μ<81.9179.29<\mu<81.91

e.

(92.82.5752.6/25)<μ<(92.8+2.5752.6/25)(92.8-2.575\cdot2.6/\sqrt{25})<\mu<(92.8+2.575\cdot2.6/\sqrt{25})

91.46<μ<94.1491.46<\mu<94.14


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