Question

5.4 Find the p-value, Part II. An independent random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given set of hypotheses and T test statistic. Also determine if the null hypothesis would be rejected at α = 0.01. (a) HA : µ > 0.5, n = 26, T = 2.485 (b) HA : µ < 3, n = 18, T = 0.5

Answer 1

Answer:

a) reject, b) accept

Step-by-step explanation:

We assume one sided test, and by looking at a t table which is one sided we can get the p values.

We have:

α = 0.01

a) µ > 0.5, n = 26, T = 2.485

we have $df = n -1$ degrees of freedom

n = 26 so $df = 25$

now we look at the table in row which index is 25 and see the closest number to $T = 2.485$ we can find.

We find that number to be exactly 0.01 which is exactly equal to the critical value α, unfortunately 0.01 is not greater than 0.01 so we have to reject the hypotesis.

b) µ < 3, n = 18, T = 0.5

we do the same, this time $df =18-1=17$, now we find that the closest value to 0.5 is 0.689, we can not find an exact value!, so what we do is say that the probability lies between the probability of the closest values to our number, looking at the table, 0.257 and 0.689 are the references we have, and their respective probabilities are 0.4 and 0.25,

so we just say that p is between 0.4 and 0.25.

Now we have that either value is greater than α, so we accept the hypotesis

Answer 2

Using the t-distribution, it is found that:

a)

The p-value is of 0.010002. Since it is greater than 0.01, the null hypothesis would not be rejected.

b)

The p-value is of 0.3117. Since it is greater than 0.01, the null hypothesis would not be rejected.

When we test if it is greater or lesser than at the alternative hypothesis, we have a one-tailed test.

• The only case for a two-tailed test is if the test if of difference.

Item a:

• First, we find the number of degrees of freedom, which is 1 less than the sample size, so 26 - 1 = 25 df.
• Using a calculator for a one-tailed test with 25 df and t = 2.485, the p-value is of 0.010002. Since it is greater than 0.01, the null hypothesis would not be rejected.

Item b:

• The number of degrees of freedom is df = 17.
• The test statistic is t = 0.5.

Using a calculator, the p-value is of 0.3117. Since it is greater than 0.01, the null hypothesis would not be rejected.

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