Consider a paint-drying situation in which drying time for a test specimen is no

Question

Consider a paint-drying situation in which drying time for a test specimen is normally distributed with = 7. The hypotheses H0: = 74 and Ha: < 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 = 0.005?
Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)

State the conclusion in the problem context.

(c) For the test procedure with = 0.005, what is (70)? (Round your answer to four decimal places.)
(70) =  

(d) If the test procedure with = 0.005 is used, what n is necessary to ensure that (70) = 0.01? (Round your answer up to the next whole number.)
n =  specimens

(e) If a level 0.01 test is used with n = 100, what is the probability of a type I error when = 76? (Round your answer to four decimal places.)

Explanation / Answer

Solution:-

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: > 74
Alternative hypothesis: < 74

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.005. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 1.4

z = (x - ) / SE

b)

z = - 1.21

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a z statistic test statistic of - 1.21.

Thus the P-value in this analysis is 0.113.

Interpret results. Since the P-value (0.113) is greater than the significance level (0.005), we cannot reject the null hypothesis.

c) For the test procedure with = 0.005, (70) = 0.39

d) If the test procedure with = 0.005 is used, then n necessary to ensure that (70) = 0.01 is 74.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.