Question

Radon exposure in Egyptian tombs. Refer to the Radiation Protection Dosimetry (December 2010) study of radon exposure in Egyptian tombs, Exercise 6.30 (p. 349). The radon levels—measured in becquerels per cubic meter (\({{Bq} \mathord{\left/ {\vphantom {{Bq} {{m^3}}}} \right. \\} {{m^3}}}\) )—in the inner chambers of a sample of 12 tombs are listed in the table shown below. For the safety of the guards and visitors, the Egypt Tourism Authority (ETA) will temporarily close the tombs if the true mean level of radon exposure in the tombs rises to 6,000\({{Bq} \mathord{\left/ {\vphantom {{Bq} {{m^3}}}} \right. \\} {{m^3}}}\) . Consequently, the ETA wants to conduct a test to determine if the true mean level of radon exposure in the tombs is less than 6,000\({{Bq} \mathord{\left/ {\vphantom {{Bq} {{m^3}}}} \right. \\} {{m^3}}}\) , using a Type I error probability of .10. An SPSS analysis of the data is shown at the bottom of the page. Specify all the elements of the test: \({H_0}\,,{H_a}\) test statistic, p-value,\(\alpha \) , and your conclusion.

50 910 180 580 7800 4000 390 12100 3400 1300 11900 110

Short answer

The null and alternative hypothesis are given by

\(\begin{aligned}{l}{H_0}:\mu = 6000\\{H_a}:\mu < 6000\end{aligned}\)

Test statistic is -1.766.

We reject the null hypothesis.

Step-by-step solution

Stating the hypothesis

The null and alternative hypothesis are given by

\(\begin{aligned}{l}{H_0}:\mu = 6000\\{H_a}:\mu < 6000\end{aligned}\)

Computing mean and standard deviation

The sample size is 12

The mean is given by

\(\begin{aligned}\bar x &= \frac{{50 + 910 + 180 + 580 + 7800 + 4000 + 390 + 12100 + 3400 + 1300 + 11900 + 110}}{{12}}\\ &= \frac{{42720}}{{12}}\\ &= 3560\end{aligned}\)

The standard deviation is given by

\(\begin{aligned}sd &= \sqrt {\frac{\begin{aligned}{l}{\left( { - 3510} \right)^2} + {\left( { - 2650} \right)^2} + {\left( { - 3380} \right)^2} + {\left( { - 2980} \right)^2} + {\left( {4240} \right)^2}{\left( { - 3170} \right)^2}\\ + {\left( {8540} \right)^2} + {\left( { - 160} \right)^2} + {\left( { - 2260} \right)^2} + {\left( {8340} \right)^2} + {\left( { - 3510} \right)^2} + {\left( { - 3450} \right)^2}\end{aligned}}{{11}}} \\ &= \sqrt {\frac{\begin{aligned}{l}12320100 + 7022500 + 11424400 + 8880400 + 17977600 + 10048900 + \\72931600 + 25600 + 5107600 + 69555600 + 12320100 + 11902500\end{aligned}}{{11}}} \\ &= \sqrt {\frac{{251837000}}{{11}}} \\ &= \sqrt {22894272.73} \\ &= 4784.79\end{aligned}\)

The mean and standard deviation are 3560 and 4784.79.

Test statistic

The test statistic is computed as

\(\begin{aligned}t &= \frac{{\bar x - \mu }}{{\frac{s}{{\sqrt n }}}}\\ &= \frac{{3560 - 6000}}{{\frac{{4784.79}}{{\sqrt {12} }}}}\\ &= \frac{{ - 2440}}{{1381.24}}\\ &= - 1.766\end{aligned}\)

Therefore, the test statistic is -1.766.

P-value

The Type-1 error probability is given.

Hence,\(\alpha = .10\)

Degrees of freedom are

\(\begin{aligned}df &= n - 1\\ &= 12 - 1\\ &= 11\end{aligned}\)

The tabulated value is -1.363

Conclusion

The calculated value is less than the tabulated value.

Therefore, we reject the null hypothesis.