A manufacturer of nickel-hydrogen batteries randomly selects 100 nickel plates f

Question

A manufacturer of nickel-hydrogen batteries randomly selects 100 nickel plates for test cells, cycles them a specified number of times, and determines that 13 of the plates have blistered. a) Does this provide compelling evidence for concluding that more than 10% of all plates blister under such circumstances? State and the appropriate hypotheses using a significance level of 0.05. b) If it is really the case that 16% of all plates blister under these circumstances and a sample size 200 used, how likely is it that the null hypothesis will not be rejected?

Explanation / Answer

I will specify the steps to solve this questions

1.
Step 1: First we willl State the Null hypothesis.

H0: p=0.1 i.e 10% of plate blister under this circumstances

Step 2: Now we willl State the alternative hypothesis.

H1: p > 0.1 i.e more than 10% of plate blister under this circumstances

Step 3: check the alpha level. In the problem it is given as 0.05

step 4: In the z table, check what is the rejection value. For z value of 0.05, the z score equals 1.645
Z(0.05) = 1.645

step 5:calculate the mean (p bar).
p(bar)=14/100=0.14

step 6: Now lets calculate the z value
z(calculated)= (0.14-0.1)/(square root ((0.1*0.9)/100))=1.33

Since z(calculated) in step 6 is less that z score given in step 4, we accept the null hypothesis

But since it is given that rejections have already happened and our test failed to reject the Null hyothesis, we have possibly committed a Type-II error

A Type II error occurs when we fail to reject the null hypothesis H0 when the alternative hypothesis H1 is true. In standard deniotios we use = P(Type II Error).

2)

This question solutions is looking for power of Beta or Type II Error rate

Step 1: State the Null hypothesis.
H0: p=0.16 (16% have plate blisters).

Step 2: State the alternative hypothesis.
H1: p>0.16 (More than 16% have plate blisters).

Step 3
Calculate Beta by substituting values in standard formulas

(0.16)=1- (z)
(z) is a standard notation that is used to depict cumulative distribution function of the standard normal random variable

A=(0.1-0.16)+(1.645*SQRT((0.16*0.84)/200))=-0.017356726
B=SQRT((0.16*0.84)/200)=0.025922963
A/B=(-0.017356726/0.025922963)=-0.669550249
(z)= cumulative distribution function (A/B)
=0.251572269
Note: i obtained cumulative distriution funtion value by applying NORMSDIST function in excel

(0.16)=0.748427731

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.