Consider the following hypotheses: H0: mean = 5 H1: mean < 5 A test is performed with a sample of size 25. The sample mean was 4.73 and the population standard deviation is 1.2. Assume that the population is approximately normal. Use the TI-84 PLUS calculator to compute the P-value. Round your answer to four decimal places (for example: 0.0138). Write only a number as your answer.

Respuesta :

Answer:

0.1406.

Step-by-step explanation:

Given : [tex]H_o:\mu = 5[/tex]

           [tex]H_1:\mu < 5[/tex]  

To Find : compute the P-value.

Solution:

Sample size = 25

n < 30

So, we will go for t-test

Now we are given that  the population standard deviation is 1.2.

[tex]\sqrt{\frac{s}{n}}=1.2[/tex] Where s is the variance

[tex]\sqrt{\frac{s}{25}}=1.2[/tex]

[tex]\sqrt{s}=6[/tex]

[tex]s = 6^2[/tex]

[tex]s = 36[/tex]

So, the variance is 36

we require sample standard deviation for t test

So, Sample standard deviation = [tex]\sqrt{\frac{s}{n-1}}[/tex]

                                                  = [tex]\sqrt{\frac{36}{25-1}}[/tex]

                                                  = [tex]1.2247[/tex]

Formula of t-test = [tex]\frac{m-\mu}{\frac{s}{\sqrt{n}}}[/tex]

Where s is the standard deviation of sample

Sample mean = 4.73

Mean = [tex]\mu = 5[/tex]

s = 1.2247

n =25

So,  [tex]t=\frac{5-4.73}{\frac{1.2247}{\sqrt{25}}}[/tex]

[tex]t=1.1023[/tex]

So, p value with respect to t-table is 0.1406.

Hence p-value is  0.1406.

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