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A manufacturer of nickel-hydrogen batteries randomly selects 100 nickel plates for test cells, cycles them a specified number of times and determines that 14 of the plates have blistered.

Does this data provide compelling evidence for concluding that more than 10% of all plates blister under such circumstances?

Use Alpha =0.10.

A. What is the parameter of interest?

B. State the null and alternative hypotheses.

C. Calculate the test statistic.

D. Find the rejection region.

E. Make a decision and interpret.

F. Find a p-value corresponding to the test and compare with your decision in E.

Respuesta :

Answer:

a) Parameter of interest [tex] p[/tex] representing the true proportion of the plates have blistered.

b) Null hypothesis:[tex]p\leq 0.1[/tex]  

Alternative hypothesis:[tex]p > 0.1[/tex]  

c) [tex]z=\frac{0.14 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=1.33[/tex]  

d) For this case we need to find a value in the normal standard distribution that accumulates 0.1 of the area in the right tail and for this case is:

[tex] z_{critc}= 1.28[/tex]

e) For this case since our calculated value is higher than the critical value 1.33>1.28 we have enough evidence to reject the null hypothesis and we can conclude that the true proportion is significantly higher than 0.1

f) [tex]p_v =P(z>1.33)=0.0917[/tex]  

If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the true proportion is higher than 0.1 or 10%

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

Part a

Parameter of interest [tex] p[/tex] representing the true proportion of the plates have blistered.

X=14 represent the number of the plates have blistered.

[tex]\hat p=\frac{14}{100}=0.14[/tex] estimated proportion of the plates have blistered.

[tex]p_o=0.1[/tex] is the value that we want to test

[tex]\alpha=0.1[/tex] represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

Part b: Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that  more than 10% of all plates blister under such circumstances.:  

Null hypothesis:[tex]p\leq 0.1[/tex]  

Alternative hypothesis:[tex]p > 0.1[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Part c: Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.14 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=1.33[/tex]  

Part d: Rejection region

For this case we need to find a value in the normal standard distribution that accumulates 0.1 of the area in the right tail and for this case is:

[tex] z_{critc}= 1.28[/tex]

Part e

For this case since our calculated value is higher than the critical value 1.33>1.28 we have enough evidence to reject the null hypothesis and we can conclude that the true proportion is significantly higher than 0.1

Part f

Since is a right taild test the p value would be:  

[tex]p_v =P(z>1.33)=0.0917[/tex]  

If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the true proportion is higher than 0.1 or 10%

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