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------------------------------------------------ The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 101.2, and the standard deviation is 5.1. We wish to test [tex]H_0:\mu=101.2[/tex] versus [tex]H_1:\mu \neq 101.2[/tex] with a sample of [tex]n=8[/tex] specimens.

Calculate the P-value if the observed statistic is:
(a) [tex]\bar{x}=96.1[/tex]
(b) [tex]\bar{x}=100.8[/tex]
(c) [tex]\bar{x}=102.1[/tex]

Round your answers to four decimal places for each part:
(a)
(b)
(c)

(a) The p-value for x = 96.1 on cement mixture is approximately 0.1188. (b) The p-value for x= 100.8 is approximately 0.4283. (c) The p-value for x= 102.1 is approximately 0.3553.

To calculate the p-values for the observed statistics in the given hypothesis test, we will use the t-distribution since the population standard deviation is not known. Here's how we can calculate the p-values for each case:

(a) For x= 96.1: The observed statistic is x= 96.1. The sample size is n = 8.

First, we need to calculate the t-score:

t = (x - μ) / (s / √n)

where μ is the population mean, s is the sample standard deviation, and n is the sample size.

Here, μ = 101.2, s = 5.1, and n = 8:

t = (96.1 - 101.2) / (5.1 / √8) ≈ -1.775

Next, we need to find the p-value associated with this t-score in a t-distribution with (n - 1) degrees of freedom. In this case, we have (8 - 1) = 7 degrees of freedom.

Using a t-table or statistical software, we can find that the p-value for t ≈ -1.775 with 7 degrees of freedom is approximately 0.1188.

Therefore, the p-value for x= 96.1 is approximately 0.1188.

(b) For x= 100.8: The observed statistic is x= 100.8. The sample size is n = 8.

Using the same steps as above, we calculate the t-score:

t = (x - μ) / (s / √n)

Here, μ = 101.2, s = 5.1, and n = 8:

t = (100.8 - 101.2) / (5.1 / √8) ≈ -0.1961

Looking up the p-value for t ≈ -0.1961 with 7 degrees of freedom, we find that it is approximately 0.4283.

Therefore, the p-value for x= 100.8 is approximately 0.4283.

(c) For x= 102.1: The observed statistic is x= 102.1. The sample size is n = 8.

Again, following the same steps, we calculate the t-score:

t = (x - μ) / (s / √n)

Here, μ = 101.2, s = 5.1, and n = 8:

t = (102.1 - 101.2) / (5.1 / √8) ≈ 0.9902

Looking up the p-value for t ≈ 0.9902 with 7 degrees of freedom, we find that it is approximately 0.3553.

Therefore, the p-value for x= 102.1 is approximately 0.3553.

Learn more about p-value from the link given below.

https://brainly.com/question/30078820

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