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Iteru [2.4K]
3 years ago
7

A country is considering raising the speed limit on a road because they claim that the mean speed of vehicles is greater than 30

miles per hour. A random sample of 15 vehicles has a mean speed of 35 miles per hour and a standard deviation of 4.7 miles per hour. At ? = .01, do you have enough evidence to supprot the country's claim?
Calculate the p-value:
Mathematics
1 answer:
kondaur [170]3 years ago
6 0

Answer:

We conclude that speed is greater than 30 miles per hour.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ =  30 miles per hour

Sample mean, \bar{x} = 35 miles per hour

Sample size, n = 15

Alpha, α = 0.01

Sample standard deviation, s = 4.7 miles per hour

First, we design the null and the alternate hypothesis

H_{0}: \mu = 30\text{ miles per hour}\\H_A: \mu > 30\text{ miles per hour}

We use one-tailed(right) t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{35 - 30}{\frac{4.7}{\sqrt{15}} } = 4.120

Now, t_{critical} \text{ at 0.01 level of significance, 14 degree of freedom } = 2.624

Since,                  

t_{stat} > t_{critical}

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis and conclude that speed is greater than 30 miles per hour.

We calculate the p-value.

P-value = 0.00052

Since p value is lower than the significance level, we reject the null hypothesis and accept the alternate hypothesis. We conclude that speed is greater than 30 miles per hour.

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Demand for Tablet Computers The quantity demanded per month, x, of a certain make of tablet computer is related to the average u
soldier1979 [14.2K]

x = f ( p ) = \frac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } } \\\\ \qquad { p ( t ) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \quad ( 0 \leq t \leq 60 ) }

Answer:

12.0 tablet computers/month

Step-by-step explanation:

The average price of the tablet 25 months from now will be:

p ( 25) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { 25 } } + 200 \\= \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \times 5 } + 200\\\\=\dfrac { 400 } { 1 + \dfrac { 5 } { 8 } } + 200\\p(25)=\dfrac { 5800 } {13}

Next, we determine the rate at which the quantity demanded changes with respect to time.

Using Chain Rule (and a calculator)

\dfrac{dx}{dt}= \dfrac{dx}{dp}\dfrac{dp}{dt}

\dfrac{dx}{dp}= \dfrac{d}{dp}\left[{ \dfrac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } }\right] =-\dfrac{100}{9}p(810,000-p^2)^{-1/2}

\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}

Therefore:

\dfrac{dx}{dt}= \left[-\dfrac{100}{9}p(810,000-p^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}\right]

Recall that at t=25, p(25)=\dfrac { 5800 } {13} \approx 446.15

Therefore:

\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009

The quantity demanded per month of the tablet computers will be changing at a rate of 12 tablet computers/month correct to 1 decimal place.

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