Question

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.

Answer 1

a

The Null hypothesis is
`H_o : p = 0.01`

The defect did not exceed 0.01

b

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

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 = (k)/(n)`

substituting values

`\r p = (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{ (p (1- p ))/(n) } }`

substituting values

`t = \frac{ 0.0125 - 0.01 }{ \sqrt{ (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_{(\alpha )/(2) } * \sqrt{(\r p (1-\r p ))/(n) }`

where
`Z_{(\alpha )/(2) }` is the critical value of
`(\alpha )/(2)` which is obtained from the z-table.The value is

`Z_{(\alpha )/(2) } = Z_{(0.05 )/(2) } = 1.96`

The reason we are obtaining critical value of
`(\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

`(\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)

So

`E = 1.96 * \sqrt{( 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