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

Answer 1

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

Answer 2

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            $H_0:\mu=70$

  Alternative Hypothesis$H_1=\mu \neq70$

b)   $t=1.042$

c)  $TDIST(1.042)=0.30310338$

d)We reject the alternative hypothesis

Step-by-step explanation:

From the question we are told that:

Population mean $\mu=70$

Sample size $n=45$

Sample mean $\=x=72.5$

Standard deviation $\sigma=16.1 pounds.$

Significance level $\alpha=0.04$

a

Generally the Hypothesis is mathematically given by

Null Hypothesis            $H_0:\mu=70$

Alternative Hypothesis$H_1=\mu \neq70$

b) Generally the Equation for test statistics is mathematically given by

  $t=\frac{\=x-\mu}{\frac{s}{\sqrt{n} } }$

  $t=\frac{72.5-70}{\frac{16.1}{\sqrt{45}}}$

  $t=1.042$

c)

Generally From T distribution table P value is mathematically given by

  $TDIST(1.042)=0.30310338$

d)

Therefore as p value is greater tab significance level

$0.30310338>0.04$

The Test statistics does nt fall in the rejection rejoin

Therefore

We reject the alternative hypothesis