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9966 [12]
2 years ago
8

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?
Mathematics
1 answer:
pshichka [43]2 years ago
6 0

Answer:

(a) The value \bar x = 72.3 is 1.21 standard deviations below the null value of <em>μ</em>.

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

Step-by-step explanation:

The hypothesis for the single mean test is:

<em>H₀</em>: <em>μ</em> = 74 vs. <em>Hₐ</em>: μ < 74.

The information provided is:

<em>σ</em> = 7

<em>n</em> = 25

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

(a)

The <em>z</em>-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 <em>z</em>-score for sample mean is given by:

z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}

If \bar x = 72.3 compute the corresponding <em>z</em>-score as follows:

z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}

  =\frac{72.3-74}{7/\sqrt{25}}\\=-1.21

Thus, the value \bar x = 72.3 is 1.21 standard deviations below the null value of <em>μ</em>.

(b)

So the test statistic is, <em>z</em> = -1.21.

Compute the <em>p</em>-value of the test as follows:

p-value=P(Z

*Use a <em>z</em>-table.

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

<em>p</em>-value = 0.1131 < <em>α</em> = 0.01

The null hypothesis was failed to be rejected.

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

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