Respuesta :
Answer:
z(s) is in the acceptance region we accept H₀. We don´t have enough evidence to support the breeder´s claim
Step-by-step explanation:
We will test the breeder´s claim at 95% ( CI) or significance level
α = 5 % α = 0,05 α /2 = 0,025
Sample Information:
sample size n = 45
sample mean x = 72,5 pounds
Sample standard deviation s = 16,1
1.-Hypothesis Test:
Null Hypothesis H₀ x = 70
Alternative Hypothesis Hₐ x ≠ 70
Alternative hypothesis contains the information about what kind of test has to be developed ( in this case it will be a two-tail tets)
2.-z (c) is from z-table z(c) = 1,96
3.- z(s) = ( x - 70 ) / 16,1 / √45
z(s) = (72,5 -70 ) *√45 / 16,1
z(s) = 2,5 * 6,71 / 16,1
z(s) = 1,04
4.-Comparing z(s) and z(c)
z(s) < z(c)
Then z(s) is in the acceptance region we accept H₀. We don´t have enough evidence to support the breeder´s claim
Complete Question:
It is assumed that the mean weight of a Labrador retriever is 70 pounds. A breeder claims that the average weight of an adult male Labrador retriever is not equal to 70 pounds. A random sample of 45 male Labradors weigh an average of 72.5 pounds with a standard deviation of 16.1 pounds. Test the breeder's claim at \alpha=0.04
a)State null and alt hypothesis
b)determine t statistics
c)compute the P value
d) decision about the test
Answer:
a)Null Hypothesis [tex]H_0:\mu=70[/tex]
Alternative Hypothesis[tex]H_1=\mu \neq70[/tex]
b) [tex]t=1.042[/tex]
c) [tex]TDIST(1.042)=0.30310338[/tex]
d)We reject the alternative hypothesis
Step-by-step explanation:
From the question we are told that:
Population mean [tex]\mu=70[/tex]
Sample size [tex]n=45[/tex]
Sample mean [tex]\=x=72.5[/tex]
Standard deviation [tex]\sigma=16.1 pounds.[/tex]
Significance level [tex]\alpha=0.04[/tex]
a
Generally the Hypothesis is mathematically given by
Null Hypothesis [tex]H_0:\mu=70[/tex]
Alternative Hypothesis[tex]H_1=\mu \neq70[/tex]
b) Generally the Equation for test statistics is mathematically given by
[tex]t=\frac{\=x-\mu}{\frac{s}{\sqrt{n} } }[/tex]
[tex]t=\frac{72.5-70}{\frac{16.1}{\sqrt{45}}}[/tex]
[tex]t=1.042[/tex]
c)
Generally From T distribution table P value is mathematically given by
[tex]TDIST(1.042)=0.30310338[/tex]
d)
Therefore as p value is greater tab significance level
[tex]0.30310338>0.04[/tex]
The Test statistics does nt fall in the rejection rejoin
Therefore
We reject the alternative hypothesis
