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V125BC [204]
3 years ago
14

The batteries produced in a manufacturing plant have a mean time to failure of 30 months, with a standard deviation of 2 months.

I select a simple random sample of 400 batteries produced in the manufacturing plant. I test each and record how long it takes for each battery to fail. I then compute that the average failure time of the 400 batteries is 29.9 months with a standard deviation of 2.15 months. In this scenario, the value 30 is:
Mathematics
1 answer:
Oduvanchick [21]3 years ago
8 0

Answer:

The population mean.

Step-by-step explanation:

29.9 is the <u>sample</u> mean.

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A plastics manufacturer has developed a new type of plastic trash can and proposes to sell them with an unconditional 6-year war
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Answer:

since -3.73 is less than 1.645, we reject H₀.    

Therefore this indicate that the proposed warranty should be modified

Step-by-step explanation:

Given that the data in the question;

p" = 13/20 = 0.65

Now  the test hypothesis;

H₀ : p = 0.9

Hₐ : p < 0.9

Now lets determine the test statistic;

Z = (p" - p ) / √[p×(1-p)/n]

= (0.65 - 0.9) /√[0.9 × (1 - 0.9) / 20]  

= -0.25 / √[0.9 × 0.1 / 20 ]

= -0.25 / √0.0045

= -0.25 / 0.067

= - 3.73

Now given that a = 0.05,

the critical value is Z(0.05) = 1.645 (form standard normal table)

Now since -3.73 is less than 1.645, we reject H₀.    

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3 0
2 years ago
A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. ​a) E
polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) \alpha =1-0.98=0.02

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq  p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and z_{\alpha/2} is the z-value that let a probability of \alpha/2 on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, \alpha by 0.02 and z_{\alpha/2} by 2.33

Where p' and \alpha are calculated as:

p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02

So, replacing the values we get:

0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq  p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence 1-\alpha is equal to 98%

It means the the level of significance \alpha is:

\alpha =1-0.98=0.02

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