P Value Lecture 31

 

P Value

Lecture 31

The probability value (P value) is a numerical measure of the strength of evidence against the null hypothesis. It is a probability at which the sample data null hypothesis is rejected, assumed to be true. If the P value is less than the significance, the null hypothesis is rejected, and the smallest value indicates strong evidence to reject a null hypothesis. If the P value is greater than the significance level, it fails to reject a null hypothesis.

When formulating null and alternative hypotheses, take into account the following three situations:.

Case 1: Left-tailed test

If H0: μ < μ0, then the critical region is at the left side of the sampling distribution of a statistic.

P value = P(z < z calculated)

Case 2: Right-tailed test

If H0: μ > μ0, then the critical region is at the right side of the sampling distribution of a statistic.

P value = P(z > z calculated)

Case 3: Two-tailed test

If H0: μ  ≠ μ0, then the critical region is at both sides of the sampling distribution of a statistic.

Reject  H0 if P <  α and fail to reject H0 if P >  α

in case of two-tailed test 

Reject  H0 if 2P <  α 

Where:

P value = P(z < z calculated) + P value = P(z > z calculated)

The p value can be determined directly using the following table:

Example 8.9: The average weight of residents in a town is 168 lb. A nutritionist believes that the true mean is to be different. She measured 36 individuals and found the mean weight to be 169.5 lbs. with a standard deviation of 3.90. Test opinion about the mean weight at 5 % level of significance, using

a. Traditional method

b. P value method

Solution: (a): Traditional method:

i.  $H_0: \mu = 168 Vs. H_1: \mu \ne 168$

$$\begin{gathered} \text{ \\ ii. The significance level; α = 0.05} \end{gathered}$$

$$\begin{gathered} \text{ \\ iii. The test statistic: } \end{gathered}$$As the sample size is large, then we using

vi. Reject H0, if |z| ≥ z0.05/2 = 1.96

v. Computation:

vi. Remarks: The calculated z value falling in the rejection area (Reject H0), it is concluded that the nutritionist believes is true.

(b): P value method:

i.  $H_0: \mu = 168 Vs. H_1: \mu \ne 168$

$ii.\; The\; significance\; level;\; \alpha \; =\; 0.05$

$iii.\; The\; test\; statistic:$As the sample size is large, then we using

iv. Computation:

v. Computation of P value:

P value = P (z > 2.31)  

P value = 1.00 - 0.9896 

P value = 0.0104

2P = 0.0208

2P < α = 0.05

vi. Remarks: The 2P value < 0.05, the null hypothesis is rejected. It is concluded that the nutritionist believes is true.

Example 8.10: A factory manufactures cars with a warranty of five years on the engine and transmission. An engineer believes that the engine or transmission will malfunction in less than 5 years. He tests a sample of 40 cars and finds the average time to be 4.8 years with a standard deviation of 0.50. Test the hypothesis at 2 % level of significance, using

a.      Traditional method

b.      P value method.

Solution:

(a): Traditional method

i.  

$$\begin{gathered} \text{ \\ ii. The significance level; α = 0.02} \end{gathered}$$

$$\begin{gathered} \text{ \\ iii. The test statistic: } \end{gathered}$$As the sample size is large, then we using

vi. Reject H0, if  z < -z0.02 = - 2.33

v. Computation:

vi. Remarks: The calculated z value falling in the rejection area (Reject H0) means that the engineer claim is true.

 b. P value Method

i.  $H_0: \mu \mathrm{\ge \; 5} Vs. H_1: \mu <\; 5$

$ii.\; The\; significance\; level;\; \alpha \; =\; 0.02$

$iii.\; The\; test\; statistic:$As the sample size is large, then we using

vi. Computation:

v.  P value = P(z < - 2.53)

P value = P(z < - 2.53)

P value = 1 - P(z < - 2.53)

P value = 1 - 0.9941

P value = 0.0059

vi. Remarks: The P value is 0.0059< 0.02, (Reject H0) it is concluded that the engineer claim is true.

Example 8.11: A researcher wishes to test the claim that the average cost of per-month tuition fees at a private college is greater than Rs. 5550. He selects a random sample of 36 students and finds the mean to be Rs. 5800. The population standard deviation is Rs. 600. Is there evidence to support the claim at the 5% significance level? Using the p-value method.

Solution:

i.  $H_0: \mu \mathrm{\le 5550} Vs. H_1: \mu >\; 5550$

$ii.\; The\; significance\; level;\; \alpha \; =\; 0.05$

$iii.\; The\; test\; statistic:$As the sample size is large, then we using

vi. Computation:

v. P value = P(z > 2.5) 

P value = P(z > 2.5) 

P value = 1-P(z > 2.5) 

P value = 1 - 0.9938

P value = 0.0062

vi. Remarks: The P value is 0.0062< 0.05 (Reject H0), it is concluded that the researcher claim is true.

Example 8.12: According to marketers, 85% of the adult population owns a mobile phone. According to one cell phone maker, the figure is actually lower. 117 of the 200 adults in the sample own a cell phone. At the 1% significance level, find out if the percentage of adults with mobile phones is less than what the marketers say.

Solution:

i.  ${\mathrm{H0:P= 0.85}}_{} Vs. {\mathrm{H1:P <}}_{} 0.85$

$ii.\; The\; significance\; level;\; \alpha \; =\; 0.01$

$iii.\; The\; test\; statistic:$As the sample size is large, then we using

iv. Computation:

As x < np0, using - 0.5

v. Computation of P value:

P value = P(z < -10.39)

P value = 1-P(z < -10.39)

P value = 1 - 0.9999

P value = 0.0001

vi. Remarks: The p value = 0.0001 is less than the significance level. The null hypothesis is rejected. It is concluded that the claim is not validated.

• Read More: Hypothesis Testing about Difference of Means & Proportions