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Y_Kistochka [10]
3 years ago
15

A neighbor paid Moshe $20 for taking care of his cat and an additional amount for watering his plants. Moshe uses the equation

Mathematics
1 answer:
alexandr1967 [171]3 years ago
6 0

Answer:

Choice A:  the amount Moshe was paid for watering the plants

Step-by-step explanation:

20 represents the amount for taking care of the cat

25 is the sum from adding the amount for taking care of the cat and the amount for watering the plants

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We wish to see if the dial indicating the oven temperature for a certain model of oven is properly calibrated. Four ovens of thi
antoniya [11.8K]

Answer:

a. Standard deviation: 4.082

Standard error: 2.041

b. The 95% confidence interval for the actual temperature is (298.5, 311.5).

Upper bound: 311.5

Lower bound: 298.5

c. Test statistic t=2.45

P-value = 0.092

d. There is no enough evidence to claim that the dial of the oven is not properly calibrated. The actual temperature does not significantly differ from 300 °F.

e. If we use a significance level of 10% (a less rigorous test, in which the null hypothesis is rejected with with less requirements), the conclusion changes and now there is enough evidence to claim that the dial is not properly calibrated.

This happens because now the P-value (0.092) is smaller than the significance level (0.10), given statististical evidence for the claim.

Step-by-step explanation:

The mean and standard deviation of the sample are:

M=\dfrac{1}{4}\sum_{i=1}^{4}(305+310+300+305)\\\\\\ M=\dfrac{1220}{4}=305

s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{4}(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{3}\cdot [(305-(305))^2+(310-(305))^2+(300-(305))^2+(305-(305))^2]}\\\\\\            s=\sqrt{\dfrac{1}{3}\cdot [(0)+(25)+(25)+(0)]}\\\\\\            s=\sqrt{\dfrac{50}{3}}=\sqrt{16.667}\\\\\\s=4.082

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=305.

The sample size is N=4.

When σ is not known, s divided by the square root of N is used as an estimate of σM (standard error):

s_M=\dfrac{s}{\sqrt{N}}=\dfrac{4.082}{\sqrt{4}}=\dfrac{4.082}{2}=2.041

The degrees of freedom for this sample size are:

df=n-1=4-1=3

The t-value for a 95% confidence interval and 3 degrees of freedom is t=3.18.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=3.18 \cdot 2.041=6.5

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 305-6.5=298.5\\\\UL=M+t \cdot s_M = 305+6.5=311.5

The 95% confidence interval for the actual temperature is (298.5, 311.5).

This is a hypothesis test for the population mean.

The claim is that the actual temperature of the oven when the dial is at 300 °F does not significantly differ from 300 °F.

Then, the null and alternative hypothesis are:

H_0: \mu=300\\\\H_a:\mu\neq 300

The significance level is 0.05.

The sample has a size n=4.

The sample mean is M=305.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=4.028.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{4.082}{\sqrt{4}}=2.041

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{305-300}{2.041}=\dfrac{5}{2.041}=2.45

The degrees of freedom for this sample size are:

df=n-1=4-1=3

This test is a two-tailed test, with 3 degrees of freedom and t=2.45, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=2\cdot P(t>2.45)=0.092

As the P-value (0.092) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the actual temperature of the oven when the dial is at 300 °F does not significantly differ from 300 °F.

If the significance level is 10%, the P-value (0.092) is smaller than the significance level (0.1) and the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the actual temperature of the oven when the dial is at 300 °C does not significantly differ from 300 °C.

5 0
3 years ago
Is 1/a rational or irrational
Alenkinab [10]
The number 1 can be classified as: a natural number, a whole number, a perfect square, a perfect cube, an integer. This is only possible because 1 is a RATIONAL number
6 0
2 years ago
Which of the following illustrates the truth value of the given conditional statement?
jeka57 [31]
<span>I'm pretty sure it's F T → F. I can't believe nobody answered this question yet. SMH </span>
6 0
3 years ago
8. Use the general formula for an exponential function, f(x) = a. b, to write a specific function given an
Doss [256]

Answer: f(x) = 61(3.5)^x

Step-by-step explanation:

This is what the variables always mean in the function:

a = initial value

b = growth factor

x = time

Then, you just plug them in! Hope this helps.

3 0
3 years ago
The triangles shown below must be congruent.
miskamm [114]

Answer:

true bc they all have equal angles if I'm wrong I'm sorry

4 0
3 years ago
Read 2 more answers
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