Question

The absolute viscosity for soybean oil is supposed to average 0.0078 Pa⋅sPa⋅s at 90 ∘C90 ∘C. Suppose a food scientist collects a random sample of 5 quantities of soybean oil and computes the mean viscosity for his sample to be ¯¯¯x=0.0083 Pa⋅sx¯=0.0083 Pa⋅s at 90 ∘C90 ∘C. Assume that measurement errors are normally distributed and that the population standard deviation of soybean oil viscosity is known to be σ=0.0005 Pa⋅sσ=0.0005 Pa⋅s.

The scientist will use a one‑sample zz‑test for a mean, at a significance level of α=0.05α=0.05, to evaluate the null hypothesis, H0:μ=0.0078 Pa⋅sH0:μ=0.0078 Pa⋅s against the alternative hypothesis, H1:μ≠0.0078 Pa⋅sH1:μ≠0.0078 Pa⋅s. Complete the scientist's analysis by calculating the value of the one-sample zz‑statistic, the pp‑value, and then deciding whether to reject the null hypothesis.

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

z=

Determine the P-valueP-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=0.05α=0.05?

a) Yes. Because the P-value of the test is greater than the stated alpha level of 0.05, there is sufficient evidence to reject the null hypothesis that the mean viscosity of soybean oil is equal to 0.0078 Pa⋅s at 90 ∘C.to 0.0078 Pa⋅s at 90 ∘C.

b) No. Because the P-value of the test is less than the stated alpha level of 0.05, there is insufficient evidence to reject the null hypothesis that the mean viscosity of soybean oil is equal to 0.0078 Pa⋅s at 90 ∘C.to 0.0078 Pa⋅s at 90 ∘C.

c) Yes. Because the P-value of the test is less than the stated alpha level of 0.05, there is sufficient evidence to reject the null hypothesis that the mean viscosity of soybean oil is equal to 0.0078 Pa⋅s at 90 ∘C.to 0.0078 Pa⋅s at 90 ∘C.

d) No. Because the P-value of the test is greater than the stated alpha level of 0.05, there is insufficient evidence to reject the null hypothesis that the mean viscosity of soybean oil is equal to 0.0078 Pa⋅s at 90 ∘C.

Answer 1

Result:

The absolute viscosity for soybean oil is supposed to average 0.0078 Pa⋅s at 90 ∘C. Suppose a food scientist collects a random sample of 5 quantities of soybean oil and computes the mean viscosity for his sample to be ¯¯¯x=0.0083 Pa⋅sx¯= at 90 ∘C. Assume that measurement errors are normally distributed and that the population standard deviation of soybean oil viscosity is known to be σ=0.0005 Pa⋅s.

The scientist will use a one‑sample zz‑test for a mean, at a significance level of α=0.05, to evaluate the null hypothesis, H0:μ=0.0078 Pa⋅ against the alternative hypothesis, H1:μ≠0.0078 Pa⋅. Complete the scientist's analysis by calculating the value of the one-sample zz‑statistic, the pp‑value, and then deciding whether to reject the null hypothesis.

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

z=2.24

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=0.0253

Should the researcher reject his null hypothesis if his significance level is α=0.05?

c) Yes. Because the P-value of the test is less than the stated alpha level of 0.05, there is sufficient evidence to reject the null hypothesis that the mean viscosity of soybean oil is equal to 0.0078 Pa⋅s at 90 ∘C.

Ho: µ = 0.0078   H1: µ ≠ 0.0078

=2.2361

Z Test of Hypothesis for the Mean
Data
Null Hypothesis                       m=	0.0078
Level of Significance	0.05
Population Standard Deviation	0.0005
Sample Size	5
Sample Mean	0.0083
Intermediate Calculations
Standard Error of the Mean	0.0002
Z Test Statistic	2.2361
Two-Tail Test
Lower Critical Value	-1.9600
Upper Critical Value	1.9600
p-Value	0.0253
Reject the null hypothesis