Mathematics, 05.05.2020 05:25 alydiale584

A sample of 39 cookies is taken to test the claim that each cookie contains, on average, 8 chocolate chips. The average number of chocolate chips per cookie in the sample was 8.2 with a standard deviation of 0.6. Assume the distribution of the population is normal. Use a 0.01 level of significance. Choose the t-statistic and the appropriate conclusion for testing this claim

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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:

$H_0: \mu = 8$

At the alternative hypothesis, it is tested if the average is different of 8, that is:

$H_1: \mu \neq 8$

We have the standard deviation for the sample, thus, the t-distribution is used. The test statistic is given by:

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

The parameters are:

$\overline{x}$ is the sample mean. $\mu$ 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:

$\overline{x} = 8.2, \mu = 8, s = 0.6, n = 39$

Hence, the value of the test statistic is:

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

$t = \frac{8.2 - 8}{\frac{0.6}{\sqrt{39}}}$

t = 2.08

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 link

A sample of 39 cookies is taken to test the claim that each cookie contains, on average, 8 chocolate

Answer from: Artemis3821

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

$t = \frac{x^{-}-mean }{\frac{S}{\sqrt{n} } }$

$t = \frac{8.2-8 }{\frac{0.6}{\sqrt{39} } }$

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.

Answer from: Quest

its kind of blurry. can you put a clearer pic?

step-by-step explanation:

Answer from: Quest

$x \times 6 = 108$

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