Question

Consider the following hypothesis test:

H0: μ ≤ 29
Ha: > 29
(a) Compute the value of the test statistic (to 2 decimals).
(b) What is the p-value (to 4 decimals)? Use the value of the test statistic rounded to 2 decimal places in your calculations.

Answer 1

a)
`z=(30.4-29)/((6)/(√(40)))=1.48`

b)
`p_v =P(z>1.48)=0.0694`

c) If we compare the p value and the significance level given
`\alpha=0.01` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis

d) For this case since we have a right tailed test we need to find a critical value on the normal standard distribution who accumulates 0.01 of the area on the right tail, and this value is:
`z_(crit)= 2.326`

And the rejection zone for H0 would be
`z>2.326`

And since our calculated value 1.48<2.326 we FAIL to reject the null hypothesis at 1% of significance.

Explanation:

The problem is complete like this:

A sample of 40 provided a sample mean of 30.4. The population standard deviation is 6.

a. Compute the value of the test statistic (to 2 decimals).

Data given and notation

`\bar X=30.4` represent the sample mean

`\sigma=6` represent the population standard deviation for the sample

`n=40` sample size

`\mu_o =29` represent the value that we want to test

`\alpha` represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 29, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 29`

Alternative hypothesis:
`\mu > 29`

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

`z=(\bar X-\mu_o)/((\sigma)/(√(n)))` (1)

z-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 can replace in formula (1) the info given like this:

`z=(30.4-29)/((6)/(√(40)))=1.48`

b. What is the p-value (to 4 decimals)? Use the value of the test statistic rounded to 2 decimal places in your calculations.

P-value

Since is a right tailed test the p value would be:

`p_v =P(z>1.48)=0.0694`

c. At = .01, what is your conclusion? p-value _____ is H0

If we compare the p value and the significance level given
`\alpha=0.01` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis

d. What is the rejection rule using the critical value? Reject H0 if z _____ is What is your conclusion? ____ H0

For this case since we have a right tailed test we need to find a critical value on the normal standard distribution who accumulates 0.01 of the area on the right tail, and this value is:
`z_(crit)= 2.326`

And the rejection zone for H0 would be
`z>2.326`

And since our calculated value 1.48<2.326 we FAIL to reject the null hypothesis at 1% of significance.