Respuesta :
Answer:
The calculated value t = 2.083 < 2.429 at 0.01 level of significance with 38 degrees of freedom.
The null hypothesis is accepted
That is x⁻ and 'μ' do not differ significantly.
Step-by-step explanation:
Step(i):-
Given sample size 'n' = 39
The mean of the Population 'μ' = 8
The mean of the sample x⁻ = 8.2
The standard deviation of the sample 'S' = 0.6
Level of significance ∝=0.01
Step(ii):-
Null hypothesis : H₀ : x⁻ = 'μ'
Alternative hypothesis: H₁: x⁻ ≠ 'μ'
Degrees of freedom ν = n-1 = 39-1 =38
t₀.₉₉ = 2.429 ( see t- table)
The test statistic
[tex]t = \frac{x^{-}-mean }{\frac{S}{\sqrt{n} } }[/tex]
[tex]t = \frac{8.2-8 }{\frac{0.6}{\sqrt{39} } }[/tex]
t = 2.083 < 2.429 at 0.01 level of significance
Step (iii):-
The calculated value t = 2.083 < 2.429 at 0.01 level of significance with 38 degrees of freedom.
The null hypothesis is accepted
That is x⁻ and 'μ' do not differ significantly.
Using the t-distribution, it is found that since the p-value of the test is of 0.0443 > 0.01, there is not enough evidence to conclude that the average is different of 8 chips.
At the null hypothesis, it is tested if the average is of 8, that is:
[tex]H_0: \mu = 8[/tex]
At the alternative hypothesis, it is tested if the average is different of 8, that is:
[tex]H_1: \mu \neq 8[/tex]
We have the standard deviation for the sample, thus, the t-distribution is used. The test statistic is given by:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
The parameters are:
- [tex]\overline{x}[/tex] is the sample mean.
- [tex]\mu[/tex] is the value tested at the null hypothesis.
- s is the standard deviation of the sample.
- n is the sample size.
For this problem, the values of the parameters are:
[tex]\overline{x} = 8.2, \mu = 8, s = 0.6, n = 39[/tex]
Hence, the value of the test statistic is:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{8.2 - 8}{\frac{0.6}{\sqrt{39}}}[/tex]
[tex]t = 2.08[/tex]
The p-value is found using a two-tailed test, as we are testing if the mean is different of a value, with t = 2.08 and 39 - 1 = 38 df.
- Using a t-distribution calculator, this p-value is of 0.0443.
Since the p-value of the test is of 0.0443 > 0.01, there is not enough evidence to conclude that the average is different of 8 chips.
A similar problem is given at https://brainly.com/question/13873630
