Question

A sample of 12 radon detectors of a certain type was selected, and each was exposed to 100 pcI/L of radon. The resulting readings were as follows: 105.5 91.4 100.1 91.7 97.0 91.8 97.4 92.9 108.2 100.6 103.8 106.1 in USE SALT (a) Does this data suggest that the population mean reading under these conditions differs from 100? (Use a = 0.05.) State the appropriate null and alternative hypotheses. Ho: 4 = 100 Haul + 100 Ho: 4 = 100 Ha: 4 < 100 #g: 4 = 100 Hus 100 Hoiu = 100 H: 4 > 100 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) P-value = State the conclusion in the problem context. Reject the null hypothesis. There is sufficient evidence to conclude that the population mean reading under these conditions differs from 100 pCi/L. Do not reject the null hypothesis. There is sufficient evidence to conclude that the population mean reading under these conditions differs from 100 pCi/L. Do not reject the null hypothesis. There is not sufficient evidence conclude that the population mean reading under these conditions differs from 100 pCi/L. Reject the null hypothesis. There is not sufficient evidence to conclude that the population mean reading under these conditions differs from 100 PCI/L. (b) Suppose that prior to the experiment a value of 0 = 7.5 had been assumed. How many determinations would then have been appropriate to obtain B = 0.10 for the alternative j = 957 (You will need to use the appropriate table in the Appendix of Tables to answer this question.) determinations

Answer 1

From the given data

Statistics Total Sum of Variable Count Mean StDev Variance Sum Squares Sample 12 98.87 6.11 37.33 1186.50 117725.77

a) Correct answer: Option (A) H0: mu = 100 and Ha: Mu not = 100

Test Statistic:

The t-statistic is computed as follows: X - MO t= 98.87 – 100 6.11/12 = -0.641 8/ſn

Test Statistic t = -0.64

The p-value is p = 0.535

Correct answer: option (C) Do not reject the null hypothesis. There is no sufficient evidence to conclude that the population mean reading under these conditions differs from 100 pCI/L

b) Answer: Sample size: 24

since

| 4-kol 95-1001 Effect Size ES = = 0.6667 U 7.5 Given a = 0.05, For 90% Power 20.90 = 1.282 (Z1-a2 +Z14 Sample size n= ES Z 0.05 +21-0.10 = 2 0.6667 1.96+1.282 0.6667 = 24 Therefore, a sample of size n = 24 will ensure that a two-sided test with a = 0.05 has 90% power to detect 5 difference in means.