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 than 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 test the appropriate hypotheses using a significance level of 0.05. If it is really the case that 16% of all plates blister under these circumstances and a sample size 200 is used, how likely is it that the null hypothesis will not be rejected?

Explanation / Answer

1]

Here we want to test that "more than 10% of all plates blisterred"

So null hypothesis is H0 : P < 0.10

and alternative hypothesis is H1 : P >= 0.10

We can used one sample proportion z test.

Let' write given information.

n = sample size = 100

x = number of favourable events = 13

level of significance = 0.05

Using minitab.

The command for one sample proportion z test in minitab is

Stat>>>Basic statistics>>>1-proportion...

Then click on summarized data

number of events = x = 13

Number of trials = n = 100

Click on "Perform hypothesis test

Hypothesized proportion = P = 0.10

then click on option

Level of confidence in percentage = c = ( 1- lpha)*100 = = (1 -0.05)*100 = 95.0

so put "Confidence level " = 95.0

Alternative = Greater than

then click on "Use test and interval based on normal approximation"

Then click on OK and again click on OK

So we get the following output

MTB > POne 100 13;
SUBC>   Test 0.10;
SUBC>   Alternative 1;
SUBC>   UseZ.

Test and CI for One Proportion

Test of p = 0.1 vs p > 0.1

                             95% Lower
Sample   X    N    Sample p      Bound Z-Value P-Value
1           13 100 0.130000 0.074683     1.00    0.159

Using the normal approximation.

Decision rule:

1) If p-value < level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value = 0.159 > 0.05 so we used 2nd rule.

That is we fail to reject null hypothesis

Conclusion: At 5% level of significance there are insufficient evidence to say that the more than 10% of all plates are blisterred.

2]

here just change sample size n = 200 rather than 100

MTB > POne 200 13;
SUBC>   Test 0.10;
SUBC>   Alternative 1;
SUBC>   UseZ.

Test and CI for One Proportion

Test of p = 0.1 vs p > 0.1

                             95% Lower
Sample   X    N   Sample p      Bound Z-Value P-Value
1         13 200 0.065000 0.036327    -1.65    0.951

Using the normal approximation.

here also, p value = 0.951 > 0.05 so we used 2nd rule.

That is we fail to reject null hypothesis

same conclusion follow as in 1]

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