Answer :
The p-value of the distribution of population standard deviation of 33.17 texts per day is 0.9435
The test statistic's "p-value" measures how likely it is that future results will be at least as extreme as the one your sample yielded. You must understand that the null hypothesis is the underlying presumption to compute this probability.
The parameters are as follows:
The sample mean ([tex]\bar x[/tex]) = 118.4
The population standard deviation ([tex]\sigma[/tex]) = 33.17
The sample size (n) = 30
Calculate the z-score first.
[tex]z = \bold{ \frac{x - \bar x}{\frac{\sigma}{\sqrt{n} } } }[/tex]
according to the entire question, x = 128
So, we have:
[tex]z = \bold{ \frac{x - \bar x}{\frac{\sigma}{\sqrt{n} } } }\\\\z = \bold{ \frac{9.6}{\frac{33.17}{\sqrt{30} } } }\\\\z = \bold{ \frac{9.6}{\frac{33.17}{5.477 } } }\\\\z = \bold{ \frac{9.6}{ 6.056} }\\\\z = 1.5852[/tex]
As determined by the z table of probability, p = 0.9435395
or p = 0.9435 (approx.)
Consequently, the distribution's p-value is 0.9435.
Click the following link to learn more about p-values:
brainly.com/question/17571541
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