Mathematics: Atrucking firm has a large inventory of spare parts that have been in storage for a long time. it knowsthat some proportion of these spare parts will have deteriorated to the point where they are no longerusable. in order to estimate this proportion, the firm tests 75 parts, and finds that 0.25 of them are notusable. find the 95% confidence interval for the population proportion of parts that are not usable. a. 0.25(1 0.25) 0.25(1 0.25) c 0.25 1.96 p 0.25 1.96 0.95 75 75 b. 0.25(1 0.25) 0.25(1 0.25)

The answer to the problem: what percent of 350 is 56 is 16 percent. you divide 56 by 350 and multiply it by 100!...

Atrucking firm has a large inventory of spare

Ответ:

anyaimartinez1901

10.01.2022

Mathematics

$0.1546\leq \widehat{p}\leq 0.3513$

Step-by-step explanation:

The firm tests 75 parts, and finds that 0.25 of them are notusable

n = 75

x = 0.25 \times 75 = 18.75≈19

$\widehat{p}=\frac{x}{n}$

$\widehat{p}=\frac{19}{75}$

$\widehat{p}=0.253$

Confidence level = 95%

So, Z_\alpha at 95% = 1.96

Formula of confidence interval of one sample proportion:

$=\widehat{p}-Z_\alpha \sqrt{\frac{\widehat{p}(1-\widehat{p}}{n}}\leq \widehat{p}\leq \widehat{p}+Z_\alpha \sqrt{\frac{\widehat{p}(1-\widehat{p}}{n}}$

$=0.253-(1.96)\sqrt{\frac{0.253(1-0.253)}{75}}\leq \widehat{p}\leq0.253+(1.96)\sqrt{\frac{0.253(1-0.253}{75}}$

Confidence interval $=0.1546\leq \widehat{p}\leq 0.3513$

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