With domestic sources of building supplies running low several years ago, roughly 60,000 homes were built with imported Chinese drywall. According to an article federal investigators identified a strong association between chemicals in the drywall and electrical problems, and there is also strong evidence of respiratory difficulties due to the emission of hydrogen sulfide gas. An extensive examination of 58 homes found that 46 had such problems. Suppose these 58 were randomly sampled from the population of all homes having Chinese drywall.
(a) Does the data provide strong evidence for concluding that more than 50% of all homes with Chinese drywall have electrical/environmental problems? Carry out a test of hypotheses using a = 0.01.
Compute the test statistic value. Round your answer to two decimal places.
(b) Calculate a lower confidence bound using a confidence level of 99% for the percentage of all such homes that have electrical/environmental problems. (Round your answer to one decimal place.)
(c) If it is actually the case that 80% of all such homes have problems, how likely is it that the test of (a) would not conclude that more than 50% do? (Round your answer to four decimal places.)

Answer:

Step-by-step explanation:

Null hypothesis:

[tex]H_o : \mu = 0.50[/tex]

Alternative hypothesis:

[tex]H_a : \mu > 0.50[/tex]

The sample proportion [tex]\hat p = \dfrac{x}{n}[/tex]

[tex]\hat p = \dfrac{46}{58}[/tex]

[tex]\hat p = 0.7931[/tex]

The test statistics:

[tex]Z = \dfrac{\hat p - p}{\sqrt{\dfrac{p(1-p)}{n} } }[/tex]

[tex]Z = \dfrac{0.7931 - 0.50}{\sqrt{\dfrac{0.5(1-0.5)}{58} } }[/tex]

[tex]Z = \dfrac{0.2931}{\sqrt{\dfrac{0.25}{58} } }[/tex]

[tex]Z = \dfrac{0.2931}{0.06565321643}[/tex]

Z = 4.46      (to two decimal places)

At 99% C.I

Level of significance is:

∝ = 1 - C.I

∝ = 1 - 0.99

∝ = 0.01

Thus, the lower confidence bound can be computed as:

[tex]= \hat p - Z_{\alpha} \sqrt{\dfrac{\hat p ( 1- \hat p )}{n} }[/tex]

Critical value of [tex]Z_{0.01/2} = 2.33[/tex]

[tex]= 0.7931 - 2.33 \sqrt{\dfrac{0.7931 (1-0.7931)}{58}}[/tex]

[tex]= 0.7931 - 2.33 (0.05319)[/tex]

= 0.7931 - 0.1239327

= 0.6691673

[tex]\simeq[/tex] 0.7    (to one decimal place)

(c)

Given:

[tex]P_o = 0.50[/tex]  and [tex]\hat p = 0.80[/tex]

Then:

[tex]\beta (0.8) = \phi \Big [ \dfrac{p_o-\hat p + Z_{\alpha} \sqrt{\dfrac{p_o(1-p_o)}{n}}}{\sqrt{\dfrac{\hat p(1-\hat p)}{n} } } \Big][/tex]

[tex]\beta (0.8) = \phi \Big [ \dfrac{0.5-0.8 + 2.33 \sqrt{\dfrac{0.5(1-0.5)}{58}}}{\sqrt{\dfrac{0.8(1-0.8)}{58} }}\Big][/tex]

[tex]= \phi \Big [ \dfrac{-0.3+ 2.33 (0.06565)}{\sqrt{0.0027586 }} \Big][/tex]

[tex]= \phi \Big [ \dfrac{-0.3+ 0.1529645}{0.052522} \Big][/tex]

[tex]= \phi \Big [ \dfrac{-0.1470355}{0.052522 } \Big][/tex]

[tex]= \phi \Big [-2.799\Big][/tex]

= 0.0026    ( to four decimal place).

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