Question

The absolute viscosity for soybean oil is supposed to average 0.0406 Pa⋅sPa⋅s at 30 ∘C30 ∘C. Suppose a food scientist collects a random sample of 3 quantities of soybean oil and computes the mean viscosity for his sample to be ?⎯⎯⎯=0.0398 Pa⋅sx¯=0.0398 Pa⋅s at 30 ∘C30 ∘C. Assume that measurement errors are normally distributed and that the population standard deviation of soybean oil viscosity is known to be ?=0.0005 Pa⋅sσ=0.0005 Pa⋅s.

The scientist will use a one‑sample ?z‑test for a mean, at a significance level of ?=0.05α=0.05, to evaluate the null hypothesis, ?0:?=0.0406 Pa⋅sH0:μ=0.0406 Pa⋅s against the alternative hypothesis, ?1:?≠0.0406 Pa⋅sH1:μ≠0.0406 Pa⋅s. Complete the scientist's analysis by calculating the value of the one-sample ?z‑statistic, the ?p‑value, and then deciding whether to reject the null hypothesis.

First, compute the ?z‑statistic, ?z. Provide your answer precise to two decimal places. Avoid rounding within calculations.

z=

Determine the ?p‑value of the test using either a table of standard normal critical values or software. Give your answer precise to four decimal places.

?=

Answer #1

Solution :

This is the two tailed test .

The null and alternative hypothesis is ,

H_{0} :
= 0.0406

H_{a} :
0.0406

Test statistic = z

= ( - ) / / n

= (0.0398 - 0.0406) / 0.0005 / 3

Test statistic = -2.77

P(z < -2.77) = 0.0028

P-value = 2 * 0.0028 = 0.0056

= 0.05

P-value >

Fail to reject the null hypothesis .

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