Consider a paint-drying situation in which drying time for a test specimen is normally distributed with σ = 7. The hypotheses H0: μ = 74 and Ha: μ < 74 are to be tested using a random sample of n = 25 observations. (a) How many standard deviations (of X) below the null value is x = 72.3? (b) If x = 72.3, what is the conclusion using a = 0.01?

(a) The value = 72.3 is 1.21 standard deviations below the null value of μ.

(b) The mean drying time of paint is not less than 74.

Step-by-step explanation:

The hypothesis for the single mean test is:

H₀: μ = 74 vs. Hₐ: μ < 74.

The information provided is:

σ = 7

n = 25

As the population standard deviation is known, we will use a z-test for single mean.

(a)

The z-score is a Normal distribution with mean 0 and variance 1. It is also defined as the number of standard deviations a raw score is from the mean.

The z-score for sample mean is given by:

If = 72.3 compute the corresponding z-score as follows:

Thus, the value = 72.3 is 1.21 standard deviations below the null value of μ.

(b)

So the test statistic is, z = -1.21.

Compute the p-value of the test as follows:

*Use a z-table.

The decision rule says to, reject the null hypothesis if the p-value is less than the significance level α = 0.01 and vice-versa.

p-value = 0.1131 < α = 0.01

The null hypothesis was failed to be rejected.

Thus, the mean drying time of paint is not less than 74.

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2

ljn

step-by-step explanation: