Use interval notation – (a, b), [a, b), (a, b], or [a, b] – to write the confide

Question

Use interval notation – (a, b), [a, b), (a, b], or [a, b] – to write the confidence interval in the next 2 exercises. In a sample of 100 ceramic pistons made for an experimental diesel engine, 18 were cracked.

1. Use the method in the textbook to construct an appropriate confidence interval for testing H0: p = 0.20 versus Ha: p < 0.20 at the 0.05 level of significance. p is the proportion (of
cracked pistons) in the population. Do you reject the null hypothesis?

2. Use prop.test() (without Yates' continuity correction) in R to construct an appropriate confidence interval for testing H0: p = 0.20 versus Ha: p < 0.20 at the 0.05 level of
significance. p is the proportion (of cracked pistons) in the population. Do you reject the null hypothesis?

Explanation / Answer

Result:

Use interval notation – (a, b), [a, b), (a, b], or [a, b] – to write the confidence interval in the next 2 exercises. In a sample of 100 ceramic pistons made for an experimental diesel engine, 18 were cracked.

1. Use the method in the textbook to construct an appropriate confidence interval for testing H0: p = 0.20 versus Ha: p < 0.20 at the 0.05 level of significance. p is the proportion (of cracked pistons) in the population. Do you reject the null hypothesis?

P=18/100=0.18

Z value for 95% = 1.645 ( one tail)

Margin of error = 1.645*sqrt(0.18*0.82/100) =0.063199

Upper limit: 0.18+0.0632 =0.2432

CI for proportion = [0, 0.2432]

Since 0.20 contained in the interval, null hypothesis is not rejected.

2. Use prop.test() (without Yates' continuity correction) in R to construct an appropriate confidence interval for testing H0: p = 0.20 versus Ha: p < 0.20 at the 0.05 level of
significance. p is the proportion (of cracked pistons) in the population. Do you reject the null hypothesis?

R code:

prop.test(18, 100, 0.20 ,alternative = c("less"), conf.level = 0.95, correct = FALSE)

R output

        1-sample proportions test without continuity correction

data: 18 out of 100, null probability 0.2

X-squared = 0.25, df = 1, p-value = 0.3085

alternative hypothesis: true p is less than 0.2

95 percent confidence interval:

0.0000000 0.2513522

sample estimates:

   p

0.18

Calculated chi square = 0.25, P=0.3085 which is > 0.05 level.

Ho is not rejected.

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