A recent survey reported that small businesses spend 24 hours a week marketing their business. A local chamber of commerce claims that small businesses in their area are not growing because these businesses are spending less than 24 hours a week on marketing. The chamber conducts a survey of 93 small businesses within their state and finds that the average amount of time spent on marketing is 23.0 hours a week. Assuming that the population standard deviation is 5.5 hours, is there sufficient evidence to support the chamber of commerce’s claim at the 0.02 level of significance?
Step 1 of 3 :
State the null and alternative hypotheses for the test. Fill in the blank below.
H0: μ=24
Ha: μ ____ 24
Step 2 of 3:
What is the test statistic?
Step 3 of 3:
Do we reject the null hypothesis? Is there sufficient or insufficient evidence?

Answers

Answer 1

Answer:

Ha: μ < 24

The test statistic is z = -1.75.

The pvalue of the test is 0.0401 > 0.02, which means that we do not reject the null hypothesis, as there is insufficient evidence.

Step-by-step explanation:

A recent survey reported that small businesses spend 24 hours a week marketing their business.

This means that the null hypothesis is:

[tex]H_0: \mu = 24[/tex]

A local chamber of commerce claims that small businesses in their area are not growing because these businesses are spending less than 24 hours a week on marketing.

This means that the alternate hypothesis is:

[tex]H_a: \mu < 24[/tex]

The test statistic is:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.

24 is tested at the null hypothesis:

This means that [tex]\mu = 24[/tex]

The chamber conducts a survey of 93 small businesses within their state and finds that the average amount of time spent on marketing is 23.0 hours a week.

This means that [tex]n = 93, X = 23[/tex]

The population standard deviation is 5.5 hours

This means that [tex]\sigma = 5.5[/tex]

Value of the test-statistic:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{23 - 24}{\frac{5.5}{\sqrt{93}}}[/tex]

[tex]z = -1.75[/tex]

The test statistic is z = -1.75.

Do we reject the null hypothesis? Is there sufficient or insufficient evidence?

The pvalue of the test is the probability of finding a sample mean below 23, which is the pvalue of z = -1.75.

Looking at the z table, z = -1.75 has a pvalue of 0.0401

The pvalue of the test is 0.0401 > 0.02, which means that we do not reject the null hypothesis, as there is insufficient evidence.


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Step-by-step explanation:

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Step-by-step explanation:

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A 1/17th scale model of a new hybrid car is tested in a wind tunnel at the same Reynolds number as that of the full-scale prototype. The drag coefficient of the model is the same as that of the prototype. Assuming the model and prototype are both tested in air, find the ratio of the drag on the scale model (Fm) to the drag on the prototype (Fp), i.e., Fm/Fp. Since the size of the model is 1/17th of the size of the prototype, the projected area of the model is (1/17)2 of the projected area of the prototype. Round your answer to the nearest tenth.

Answers

Answer:

The ratio of the drag coefficients [tex]\dfrac{F_m}{F_p}[/tex] is approximately 0.0002

Step-by-step explanation:

The given Reynolds number of the model = The Reynolds number of the prototype

The drag coefficient of the model, [tex]c_{m}[/tex] = The drag coefficient of the prototype, [tex]c_{p}[/tex]

The medium of the test for the model, [tex]\rho_m[/tex] = The medium of the test for the prototype, [tex]\rho_p[/tex]

The drag force is given as follows;

[tex]F_D = C_D \times A \times \dfrac{\rho \cdot V^2}{2}[/tex]

We have;

[tex]L_p = \dfrac{\rho _p}{\rho _m} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_m} \right)^2 \times L_m[/tex]

Therefore;

[tex]\dfrac{L_p}{L_m} = \dfrac{\rho _p}{\rho _m} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_m} \right)^2[/tex]

[tex]\dfrac{L_p}{L_m} =\dfrac{17}{1}[/tex]

[tex]\therefore \dfrac{L_p}{L_m} = \dfrac{17}{1} =\dfrac{\rho _p}{\rho _p} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_p} \right)^2 = \left(\dfrac{V_p}{V_m} \right)^2[/tex]

[tex]\dfrac{17}{1} = \left(\dfrac{V_p}{V_m} \right)^2[/tex]

[tex]\dfrac{F_p}{F_m} = \dfrac{c_p \times A_p \times \dfrac{\rho_p \cdot V_p^2}{2}}{c_m \times A_m \times \dfrac{\rho_m \cdot V_m^2}{2}} = \dfrac{A_p}{A_m} \times \dfrac{V_p^2}{V_m^2}[/tex]

[tex]\dfrac{A_m}{A_p} = \left( \dfrac{1}{17} \right)^2[/tex]

[tex]\dfrac{F_p}{F_m} = \dfrac{A_p}{A_m} \times \dfrac{V_p^2}{V_m^2}= \left (\dfrac{17}{1} \right)^2 \times \left( \left\dfrac{17}{1} \right) = 17^3[/tex]

[tex]\dfrac{F_m}{F_p} = \left( \left\dfrac{1}{17} \right)^3[/tex]= (1/17)^3 ≈ 0.0002

The ratio of the drag coefficients [tex]\dfrac{F_m}{F_p}[/tex] ≈ 0.0002.

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Answers

Answer:

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Step-by-step explanation:

Confidence interval concepts:

A confidence interval has two bounds, a lower bound and an upper bound.

A confidence interval is symmetric, which means that the point estimate used is the mid point between these two bounds, that is, the mean of the two bounds.

In this question:

Lower bound: 28.4

Upper bound: 38.8

Mean

(28.4+38.8)/2 = 33.6.

The mean is of 33.6

Margin of error:

(38.8 - 28.4)/2 = 5.2

The margin of error is of 5.2.

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

given

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Step-by-step explanation:

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Answers

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Answers

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Step-by-step explanation:

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Answers

Answer:

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