Mathematics, 01.08.2020 22:01 jonathanjenkins701

15. A manufacturer of electronic calculators is interested in estimating the fraction of defective units produced. A random sample of 800 calculators contains 10 defectives. a. Formulate and test the hypothesis to determine if the fraction defective exceeds 0.01. Use 0.05 significance level. b. Calculate a 95% CI for this problem. Does the CI agreed with your result on (a) explain.

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Answer from: djmccarter

The Null hypothesis is H_o : p = 0.01

The defect did not exceed 0.01

The 95% confidence interval is 0.004801 < p < 0.020199

Yes the CI agrees with the result in a because the value 0.01 fall within the CI

Step-by-step explanation:

From the question we are told that

The sample size is n = 800

The number of defective calculators is k = 10

The population is p = 0.01

The Null hypothesis is H_o : p = 0.01

The Alternative hypothesis is H_a : P 0.01

Generally the proportion of defective calculators is mathematically represented as

$\r p = \frac{k}{n}$

substituting values

$\r p = \frac{10}{800}$

$\r p = 0.0125$

Next is to obtain the critical value of $\alpha$ from the z-table.The value is

$Z_{\alpha } = 1.645$

Now the test statistics is mathematically evaluated as

$t = \frac{\r p - p }{ \sqrt{ \frac{p (1- p )}{n} } }$

substituting values

$t = \frac{ 0.0125 - 0.01 }{ \sqrt{ \frac{0.01 (1- 0.01 )}{800} } }$

t = 0.71067

Now comparing the values of t to the value of $Z_{\alpha }$ we see that $t < Z_{\alpha }$ hence we fail to reject the null hypothesis

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\r p (1-\r p )}{n} }$

where $Z_{\frac{\alpha }{2} }$ is the critical value of $\frac{\alpha }{2}$ which is obtained from the z-table.The value is

$Z_{\frac{\alpha }{2} } = Z_{\frac{0.05 }{2} } = 1.96$

The reason we are obtaining critical value of $\frac{\alpha }{2}$ instead of $\alpha$ is because $\alpha$

represents the area under the normal curve where the confidence level interval ( $1- \alpha$ ) did not cover which include both the left and right tail while

$\frac{\alpha }{2}$ is just the area of one tail which what we required to calculate the margin of error .

NOTE: We can also obtain the value using critical value calculator (math dot armstrong dot edu)

$E = 1.96 * \sqrt{\frac{ 0.0125 (1-0.0125 )}{800} }$

E = 0.007699

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p - E$

substituting values

0.0125 - 0.007699 < p < 0.0125 + 0.007699

0.004801 < p < 0.020199

Now given the p = 0.01 is within this interval then the CI agrees with answer gotten in a

Answer from: Quest

Hello!
your answer should be c

Answer from: Quest

its adding 5.5 each time

step-by-step explanation:

Will reward brainliest 15 ! need answer asap plsss

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