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-BARSIC- [3]
3 years ago
6

The weights of the watermelons grown on a farm are normally distributed. The mean weight has been μ0=32.5 pounds. Because of the

use of a new fertilizer, the farmer would like to know if the mean weight has changed. A random sample of n=9 watermelons is selected. The sample mean is x=33.8 pounds and the sample standard deviation is s=2.45 pounds. Test if the population mean has changed at a significance level a=0.05.
Mathematics
1 answer:
9966 [12]3 years ago
8 0

Answer:

t=\frac{33.8-32.5}{\frac{2.5}{\sqrt{9}}}=1.56  

p_v =2*P(z>1.56)=0.157  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can conclude that the true mean of weights for thr watermellons is not different from 32.5  5% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=33.8 represent the sample mean

s=2.45 represent the sample standard deviation

n=9 sample size  

\mu_o =32.5 represent the value that we want to test  

\alpha=0.05 represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 32.5, the system of hypothesis would be:  

Null hypothesis:\mu =32.5  

Alternative hypothesis:\mu \neq 32.5  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{33.8-32.5}{\frac{2.5}{\sqrt{9}}}=1.56  

P-value  

The degrees of freedom are given by df = n-1= 9-1=8

Since is a two-sided test the p value would be:  

p_v =2*P(z>1.56)=0.157  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can conclude that the true mean of weights for thr watermellons is not different from 32.5  5% of signficance.  

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Solve X

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W=0.0260416v

V=38.4W

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3 years ago
The distance between the points (12,9) and (0,4) is A) 14 units B) 13 units C) 15 units D) 16 units
Juliette [100K]

Answer:

Option B  13 units is correct.

Step-by-step explanation:

The formula used to find distance between 2 points is:

d(P1,P2)=\sqrt{(x_{2} -x_{1})^2 + (y_{2} -y_{1})^2 }

x₁ = 12 x₂ = 0 y₁= 9 and y₂=4

Putting values in the formula:

=\sqrt{(0-12)^2+(4-9)^2}\\=\sqrt{(-12)^2+(-5)^2}\\=\sqrt{144+25} \\=\sqrt{169} \\=13

So, Option B 13 units is correct.

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3 years ago
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Answer:

c

Step-by-step explanation:

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3 years ago
Three airlines serve a small town in Ohio. Airline A has 52% of all scheduled flights, airline B has 27% and airline C has the r
julia-pushkina [17]

Answer:

Step-by-step explanation:

Given that three airlines serve a small town in Ohio. Airline A has 52% of all scheduled flights, airline B has 27% and airline C has the remaining 21%.

Their on-time rates are 83%, 66%, and 36%, respectively.

Airlines A B C  

   

Flights                52%                  27%             21%   100%

On time rates  83%                  66%         36%  

   

Flight*online rates 43.1600% 17.8200% 7.5600% 68.5400%

Thus the above table shows the product of no of flights and on time rates.

Prob it was airline A given it left on time

= \frac{43.16}{68.54} \\=0.6297

4 0
3 years ago
The following observations are on stopping distance (ft) of a particular truck at 20 mph under specified experi- mental conditio
zzz [600]

Answer:

see explaination

Step-by-step explanation:

Data : 32.1 , 30.6 , 31.4 , 30.4 , 31.0 , 31.9

Mean X-bar = 31.23

SD = 0.689

a)

Null Hypothesis : Xbar = mu

Alternate Hypothesis : Xbar > mu

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 30 ) /(0.689/sqrt(6))

= 4.372

P-value = ~0

Since P-vale < 0.01 , we will reject null hypothesis.

The data suggest that true average stopping ditance exceeds the maximum.

b)

i) SD = 0.65

mu = 31

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 31) /(0.65/sqrt(6))

= 0.867

P-value = 0.3859 Answer

ii) SD = 0.65

mu = 32

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 32) /(0.65/sqrt(6))

= -2.9

P-value = 0.0037 Answer

c)

i) SD = 0.8

mu = 31

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 31) /(0.8/sqrt(6))

= 0.704

P-value = 0.4814 Answer

ii) SD = 0.8

mu = 32

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 32) /(0.8/sqrt(6))

= -2.357

P-value = 0.0184 Answer

The probabilities obtained in part c are comparatively higher than that of part b.

d)

i) For alpha =0.01

z = (Xbar - mu) / (SD/sqrt(n))

=> -2.32 = (31.23 - 31) /(0.65/sqrt(n))

=> (0.65/sqrt(n)) = (31.23 - 31)/-2.32

=> sqrt(n) = (0.65*(-2.32)) / (31.23 - 31)

=> n = 43 Answer

ii) For beta =0.10

z = (Xbar - mu) / (SD/sqrt(n))

=> -1.28 = (31.23 - 31) /(0.65/sqrt(n))

=> (0.65/sqrt(n)) = (31.23 - 31)/-1.28

=> sqrt(n) = (0.65*(-1.28)) / (31.23 - 31)

=> n = 13 Answer

4 0
2 years ago
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