Question

A one­sided test of a hypothesis, based on a sample of size 9, yields a P­value of .035. Which of the following best describes the possible range of t values that yields this P­value? .706 < t < .889 1.11 < t < 1.40 1.34 < t < 1.44 2.45 < t < 2.90 1.86 < t < 2.31

Answer 1

1.86 < t < 2.31

Explanation:

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean height actually is higher\lower than an specified value, the system of hypothesis would be:

Null hypothesis:
`\mu \leq \mu_o`

Alternative hypothesis:
`\mu > \mu_o`

Or like this:

Null hypothesis:
`\mu \geq \mu_o`

Alternative hypothesis:
`\mu < \mu_o`

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We don't know on this case the calculated value and with the p value we can find it.

P-value

The first step is calculate the degrees of freedom, on this case:

`df=n-1=9-1=8`

Since is a one side test the p value would be:

`p_v =P(t_((8))>t_o)=0.035`

And we can find the critical value with the following excel code:

" =T.INV(1-0.035,8)" and we got
`t_p =2.09`

And cannot be a lower tail test since all the options have positive values for the statistic. So on this case the best option is:

1.86 < t < 2.31