Question

My Netes Ask Your Consider a paint-drying situation in which drying time for a test specimen is normally distributed with ? 6. The hypotheses Ho: 74 and Hai < 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? (Round your answer to two decimal places.) standard deviations (b) If x-72.3, what is the conclusion using a-0.01? (Round your answers to two decimal places.) test stabistk What can you conclude? o Do not reject the null hypothesis. There is not sufficient evidence O Reject the null hypothesis. There is not sufficient evidence to O Reject the null hypothesis. There is sufficient evidence to conclude Do not reject the null hypothesis. There is sufficient evidence to to conclude that the mean drying time is less than 74 conclude that the mean drying time is less than 74 that the mean drying time is less than 74 conclude that the mean drying time is less than 74 (c) what is ? for the test procedure that rejects Ho when s-2547 (Round your answer to four decimal places.) (d) For the test procedure of part (c), what is B(70)? (Round your answer to four decimal places.) P(70) (e) If the test procedure of part (c) is used, what n is necessary to ensure that P(70) 0.01? (Round your answer up to the next whole number.) specimens You may need to use the appropriate table in the Appendix of Tables to answer this question

Answer 1

(a)

The z-score for $\bar{x}=72.3$ is

$z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\frac{72.3-74}{6/\sqrt{25}}=-1.42$

Answer: 1.42 standard deviations

(b)

The critical value of z for which we will reject the null hypothesis is -2.33.

Rejection region:

If z <-2.33, Reject H0

Since z is not in rejection region so we fail to reject the null hypothesis.

(C)

The $\alpha$ is

$\alpha=P(z\leq -2.54)=0.0055$

(d)

The critical value of sample mean for rejection region is

$-2.54=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\frac{\bar{x}_{c}-74}{6/\sqrt{25}}$

$\bar{x}_{c}=70.952$

For $\bar{x}_{c}=70.952$ and the z-score is

μ=70

70.952 - 7070 6/ V 25 0.79 6/V25

The type II error is

$\beta=P(z>0.79)=0.2148$

(e)

Test is left tailed so for $\alpha=P(z\leq -2.54)=0.0055$ we have $z_{1-\alpha}=2.54$

and

4-3 = 2.33

and

$\sigma=6,\delta=74-70=4$

So sample size is

$n=\left [ \frac{(z_{1-\alpha/2}+z_{1-\beta})\sigma}{\delta} \right ]^{2}=\left [ \frac{(2.54+2.33)\cdot 6}{4} \right ]^{2}=53.363025\approx 54$