The absolute viscosity for sunflower oil is supposed to average 0.0311 Pa⋅s at 38 ∘C. Suppose a food scientist collects a random sample of 4 quantities of sunflower oil and computes the mean viscosity for his sample to be x¯=0.0318 Pa⋅s at 38 ∘C. Assume that measurement errors are normally distributed and that the population standard deviation of sunflower oil viscosity is known to be σ=0.0009 Pa⋅s. The scientist will use a one‑sample z‑test for a mean, at a significance level of α=0.01, to evaluate the null hypothesis, H0:μ=0.0311 Pa⋅s against the alternative hypothesis, H1:μ≠0.0311 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. P= Should the researcher reject his null hypothesis if his significance level is α=0.01? A) Yes. Because the P-value of the test is less than the stated alpha level of 0.01, there is sufficient evidence to reject the null hypothesis that the mean viscosity of sunflower oil is equal to 0.0311 Pa⋅s at 38 ∘C. B) Yes. Because the P-value of the test is greater than the stated alpha level of 0.01, there is sufficient evidence to reject the null hypothesis that the mean viscosity of sunflower oil is equal to 0.0311 Pa⋅s at 38 ∘C. C) No. Because the P-value of the test is less than the stated alpha level of 0.01, there is insufficient evidence to reject the null hypothesis that the mean viscosity of sunflower oil is equal to 0.0311 Pa⋅s at 38 ∘C. D) No. Because the P-value of the test is greater than the stated alpha level of 0.01, there is insufficient evidence to reject the null hypothesis that the mean viscosity of sunflower oil is equal to 0.0311 Pa⋅s at 38 ∘C.
The absolute viscosity for sunflower oil is supposed to average 0.0311 Pa⋅s at 38 ∘C. Suppose a food scientist collects a random sample of 4 quantities of sunflower oil and computes the
The scientist will use a one‑sample z‑test for a mean, at a significance level of α=0.01, to evaluate the null hypothesis, H0:μ=0.0311 Pa⋅s against the alternative hypothesis, H1:μ≠0.0311 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.
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