A publisher reports that 40% of their readers own a particular make of car. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 200 found that 31% of the readers owned a particular make of car. Determine the P-value of the test statistic. Round your answer to four decimal places.

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MrRoyal MrRoyal · Answer 1 · 2021-02-20T11:45:06+01:00

Answer:

[tex]p-value = 0.9953[/tex]

Step-by-step explanation:

Given

[tex]P = 40\%[/tex] --- proportion of population

[tex]n = 200[/tex] -- samples

[tex]p = 31\%[/tex] -- proportion of samples

Required

Determine the p value

First, calculate the standard deviation ([tex]\sigma[/tex])

[tex]\sigma = \sqrt{\frac{P*(1 - P)}{n}}[/tex]

[tex]\sigma = \sqrt{\frac{40\%*(1 - 40\%)}{200}}[/tex]

[tex]\sigma = \sqrt{\frac{0.40*(1 - 0.40)}{200}}[/tex]

[tex]\sigma = \sqrt{\frac{0.40*0.60}{200}}[/tex]

[tex]\sigma = \sqrt{\frac{0.24}{200}}[/tex]

[tex]\sigma = \sqrt{0.0012}[/tex]

[tex]\sigma = 0.0346[/tex]

Next, we calculate the z score

[tex]z = \frac{p - P}{\sigma}[/tex]

[tex]z = \frac{40\% - 31\%}{0.0346}[/tex]

[tex]z = \frac{9\%}{0.0346}[/tex]

[tex]z = \frac{0.09}{0.0346}[/tex]

[tex]z = 2.6012[/tex]

[tex]z \approx 2.60[/tex]

From the z table, the p value of [tex]z = 2.60[/tex] is: 0.9953

Hence:

[tex]p-value = 0.9953[/tex]