Respuesta :
A P-value of 0.0086 is sufficient evidence to refute the manufacturer's claim.
Suppose a particular vehicle gets 50 miles per gallon on the freeway and a simple random sample of 30 cars.
Let μ indicates the average number of miles per gallon of highway for a particular car model
Let [tex]\bar{x}[/tex] indicates the sample mean, σ indicates the population standard deviation and n indicates the sample size.
Given; [tex]\bar{x}[/tex]=49, σ= 2.3 and n = 30,μ₀ = 50
The significance level is α= 5%
Hypotheses: To test the null hypothesis: H₀: μ₀ = 50 Alternative hypothesis: H₁: μ <50
The test statistic can be written as:
[tex]z=\frac{\sqrt{n}(\bar{x}-50)}{\sigma}[/tex]
It follows a standard normal distribution under H₀.
Decision rule / rejection range: P-value <0.05 or because there is a test on the left
z <Ф⁻¹ (0.05) d. H. Z <-1.6449d. H.
[tex]\bar{x}[/tex] < μ₀+(σ÷√n )×-1.6449 = 49.3092725599558
Test statistic: Observations value of the test statistic,
[tex]\begin{aligned}Z_{stat}&=\frac{\sqrt{n}(\bar{x}-50)}{\sigma}\\ Z_{stat}&=\frac{\sqrt{30}(49-50)}{2.3}\\ Z_{stat}&=-2.3814\end[/tex]
critical value = Z critical = Z0.05 = -1.6449
P-value= P(Z < -2.3814)
P-value= P(Z < -2.3814)
P-value= (–2.3814)
P-value= 0.0086
Therefore, p-value = 0.0086 <0.05 and test statistic observations Zstat critical value [tex]\nleq[/tex] -1.6449, there is enough evidence to reject the null hypothesis.
Learn more about standard normal distribution from here brainly.com/question/22872268
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