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 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 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 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.
Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.