uppose a sample of thirty 25- to 34-year-olds showed a sample mean of 118.1 texts per day. Assume a population standard deviation of 33.17 texts per day. Compute the p-value. (Round your answer to four decimal places.)

suppose a sample of thirty 25- to 34-year-olds showed a sample mean of 118.1 texts per day. Assume a population standard deviation of 33.17 texts per day. Compute the p-value if they send 128 text per day. (Round your answer to four decimal places.)

P-value or probability value is the probability of observing test results at least as extreme as the results actually measured during the test, assuming that the null hypothesis is correct.

Given that:

n = 30

mean (μ) = 118.1 texts per day

standard deviation (σ) = 33.17 texts per day

The z score (z) is given as:

[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n} } }=\frac{128-118.1}{\frac{33.17}{\sqrt{30} } }=1.63[/tex]

From the z table, P- value = P(z < 1.63) = 0.9484 = 94.84%

Therefore the P- value is 94.84%

MrRoyal MrRoyal · Answer 2 · 2021-11-16T13:27:47+01:00

The p-value of a distribution can be calculated from its z-score, and vice versa.

The p-value is 0.9489

The given parameters are:

[tex]\mathbf{\bar x= 118.1}[/tex] --- the sample mean

[tex]\mathbf{\sigma = 33.17}[/tex] --- the population standard deviation

[tex]\mathbf{n = 30}[/tex] --- the sample size

Start by calculating the z-score

[tex]\mathbf{z = \frac{x - \bar x}{\sigma /\sqrt n}}[/tex]

From the complete question, x = 128.

So, we have:

[tex]\mathbf{z = \frac{128 - 118.1}{33.17/\sqrt {30}}}[/tex]

[tex]\mathbf{z = \frac{9.9}{6.056}}[/tex]

[tex]\mathbf{z = 1.6347}[/tex]

From z table of probabilities, we have:

[tex]\mathbf{p = 0.948944}[/tex]

Approximate

[tex]\mathbf{p = 0.9489}[/tex]

Hence, the p-value is 0.9489