Question

multiple choice. choose the one alternative that best completes the statement or answers the question.
ind the p - value for the indicated hypothesis test.

1. in a sample of 88 children selected randomly from one town, it is found that 8 of them suffer from asthma. find the p - value for a test of the claim that the proportion of all children in the town who suffer from asthma is equal to 11%.

a) - 0.2843 b) 0.2843 c) 0.5686 d) 0.2157

1. a manufacturer claims that fewer than 6% of its fax machines are defective. in a random sample of 97 such fax machines, 5% are defective. find the p - value for a test of the manufacturers claim.

a) 0.3264 b) 0.1591 c) 0.3409 d) 0.1736

1. a nationwide study of american homeowners revealed that 65% have one or more lawn mowers. a lawn equipment manufacturer, located in omaha, feels the estimate is too low for households in omaha. find the p - value for a test of the claim that the proportion with lawn mowers in omaha is higher than 65%. among 497 randomly selected homes in omaha, 340 had one or more lawn mowers.

a) 0.1118 b) 0.0252 c) 0.0505 d) 0.0559

1. a random sample of 139 forty - year - old men contains 26% smokers. find the p - value for a test of the claim that the percentage of forty - year - old men that smoke is 22%.

a) 0.2542 b) 0.1271 c) 0.2802 d) 0.1401

Explanation:

Step1: Calculate sample proportion

For problem 1:
Sample proportion \( \hat{p}=\frac{8}{88}\approx0.0909\)
Population proportion \(p = 0.11\)
Sample size \(n = 88\)
Test statistic \(z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}=\frac{0.0909 - 0.11}{\sqrt{\frac{0.11\times(1 - 0.11)}{88}}}\approx - 0.56\)
Since it is a two - tailed test (claim \(p = 0.11\)), \(P - value=2\times P(Z\lt - 0.56)\)
Using standard normal table \(P(Z\lt - 0.56)=0.2877\), \(P - value\approx2\times0.2877 = 0.5754\approx0.5686\) (due to rounding differences in calculation steps)

For problem 2:
Sample proportion \( \hat{p}=0.05\)
Population proportion \(p = 0.06\)
Sample size \(n = 97\)
Test statistic \(z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}=\frac{0.05 - 0.06}{\sqrt{\frac{0.06\times(1 - 0.06)}{97}}}\approx - 0.4 \)
Since it is a left - tailed test (claim \(p\lt0.06\)), \(P - value = P(Z\lt - 0 .4)\)
Using standard normal table \(P(Z\lt - 0.4)=0.34 46\approx0.340 9\) (due to rounding differences in calculation steps)

For problem 3:
Sample proportion \( \hat{p}=\frac {340}{4 97}\approx0.6 84\) ，Population proportion \(p = 0.65\)
Sample size \(n = 497\)
Test statistic \(z=\frac{\hat {p}-p}{\sqrt{\ \frac{p(1 - p)}{n}}}=\frac{0.684 - 0.65}{\sqrt{\frac{0.65\times(1 - 0.65)}{497}}}\approx1.59\)
Since it is a right - tailed test (claim \(p\gt0.65\)), \(P - value=P(Z\gt1.59)=1 - P(Z\leqslant1.59)\)
Using standard normal table \(P(Z\leqslant1.59)=0.9441\), \(P - value=1 - 0.9441=0.0559\)

For problem 4:
Sample proportion \( \hat{p}=0.26\)
Population proportion \(p = 0.22\)
Sample size \( 13 9\)
Test statistic \(z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}=\frac{0.26 - 0.22}{\sqrt{\frac{0.22\times(1 - 0.22)}{139}}}\approx1.12\)
Since it is a two - tailed test (claim \(p = 0.22\)), \(P - value=2\times(1 - P(Z\leqslant1.12))\)
Using standard normal table \(P(Z\leqslant1.12)=0.8686\), \(P - value=2\times(1 - 0.8686)=0.2628\approx0.2542\) (due to rounding differences in calculation steps)

Answer:

1. C. 0.5686
2. C. 0.3409
3. D. 0.0559
4. A. 0.2542