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iVinArrow [24]
3 years ago
12

Construct a 90% confidence interval for μ1-μ2 with the sample statistics for mean calorie content of two​ bakeries' specialty pi

es and confidence interval construction formula below. Assume the populations are approximately normal with equal variances.

Mathematics
1 answer:
DIA [1.3K]3 years ago
4 0

Answer:

The 90% confidence interval for the difference in mean (μ₁ - μ₂) for the two bakeries is; (<u>49</u>) < μ₁ - μ₂ < (<u>289)</u>

Step-by-step explanation:

The given data are;

Bakery A

\overline x_1<em> </em>= 1,880 cal

s₁ = 148 cal

n₁ = 10

Bakery B

\overline x_2<em> </em>= 1,711 cal

s₂ = 192 cal

n₂ = 10

\left (\bar{x}_1-\bar{x}_{2}  \right ) - t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}< \mu _{1}-\mu _{2}< \left (\bar{x}_1-\bar{x}_{2}  \right ) + t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}

df = n₁ + n₂ - 2

∴ df = 10 + 18 - 2 = 26

From the t-table, we have, for two tails, t_c = 1.706

\hat{\sigma} =\sqrt{\dfrac{\left ( n_{1}-1 \right )\cdot s_{1}^{2} +\left ( n_{2}-1 \right )\cdot s_{2}^{2}}{n_{1}+n_{2}-2}}

\hat{\sigma} =\sqrt{\dfrac{\left ( 10-1 \right )\cdot 148^{2} +\left ( 18-1 \right )\cdot 192^{2}}{10+18-2}}= 178.004321469

\hat \sigma ≈ 178

Therefore, we get;

\left (1,880-1,711  \right ) - 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}< \mu _{1}-\mu _{2}< \left (1,880-1,711  \right ) + 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}

Which gives;

169 - \dfrac{75917\cdot \sqrt{35} }{3,750} < \mu _{1}-\mu _{2}< 169 + \dfrac{75917\cdot \sqrt{35} }{3,750}

Therefore, by rounding to the nearest integer, we have;

The 90% C.I. ≈ 49 < μ₁ - μ₂ < 289

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Answer: When a ball is dropped into sand, a circular crater is formed due to the impact of the ball.

The size of the crater varies with the kinetic energy of the ball at impact. In this research,

the relationship between the kinetic energy of a modified table tennis ball, dropped into

sand from a fixed height, and the diameter of the crater formed will be investigated.

The size of a meteor crater is related to the amount of energy the meteor possesses as it

strikes the ground. The more energy it possess, the larger the crater diameter. According

to the Palaeontology Department, University of Bristol website, the relationship between

the kinetic energy of a meteor at impact and diameter of the resulting crater on earth is:

4.3/1)(07.0

t

a ECfD

ρ

ρ = ⋅ ,

where D is the crater diameter, Cf is the crater collapse factor, ρa is the density of the

projectile, ρt is the density of the target rock, and E is the total energy of the meteor.

(http://palaeo.gly.bris.ac.uk/communication/Brana/impact.html)

Equation 1 describes craters formed by meteors that collide at high speed with the surface

of the earth. Given that Cf and ρt are constant (and unknown) in our research, and that

both E and ρa are proportional to the mass, m, of our projectile, equation 1 can be

simplified to the form:

4.3/1 ⋅= mEkD )( (2)

where k and n are unknown constants. For the purpose of our research, E is defined as

the kinetic energy at the instant the ball impacts the sand, and since the ball is being

dropped from constant height, it is assumed that the ball is impacting with constant

velocity. It follows that:

59.0

59.0

4.3/2

4.3/12 )5.0(

D k E

ED

ED

mvmkD

⋅=

∝

∝

⋅⋅⋅=

According to equation 3, it is predicted that the diameter of a meteor crater is

proportional to the energy of the meteor raised to the power of 0.59. It is not known if

the relationship based on meteor craters given in equation 3 will be applicable to our

situation, which involves table tennis balls filled with lead shot and latex glue dropped

into sand at low speeds. It is expected that the relationship between the kinetic energy of

the projectile and the resulting crater diameter will be a power law, but with exponent

different from that predicted by equation 3.

This research will attempt to determine the relationship between the kinetic energy at

impact of a modified table tennis ball dropped from a fixed height into sand, and the

diameter of the crater formed. It will also be determined whether it is the same as the

relationship for meteor craters.

Seven table tennis balls of the same volume and different masses were prepared as

follows. Tiny holes were drilled in each ball to insert lead shot and glue. Different

amounts of lead shot were put in to each ball to vary the mass of the balls. Latex glue

mixed with a small amount of water was inserted through the hole of the ball using a

syringe until the ball was completely filled. Glue was used to spread the lead shots evenly

inside the ball to make its density uniform. After filling each ball with glue and lead shot,

clay was used to cover the hole. A minimal amount of clay was used in order not to

change the spherical shape of the balls. The balls were massed using an electronic mass

balance.

As seen above in figure 1, a basin filled with a large amount of sand was prepared. A

circular basin was prepared so that any wall effect stays constant during the research. The

sand was filtered using a sieve with small holes. After that the sand was poured into a

basin which was placed next to a table. On the table, a stand was placed. As seen in

figure 2, three lengths were measured; L1 (top of stand to the floor), L2 (height of basin),

and L3 (top of basin to sand).

Before dropping the balls, the sand was loosened by mixing the sand with the hands. The

sand was flattened by running a straight ruler across the top of the loosened sand. Each

ball was dropped from the top of the stand (see figure 1). The ball was dropped using two

fingers. Each finger was released simultaneously so that the ball did not spin as it fell.

After the ball impacted the sand, the diameter of the crater was measured using a ruler.

The diameter of the outer edge of the crater was measured as seen in figure 3.

After measuring the diameter of

the crater, the sand was loosened

again and the same procedure

was repeated. For each ball, a

total of 5 trials was conducted to

quantify random errors.

Throughout the experiment, the

same type of sand was used, and

the whole experiment was

conducted in the laboratory room

in one hour.

Step-by-step explanation: There is more to tell but this is all for now

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