A sample of 1700 computer chips revealed that 35% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature claimed that over 32% do not fail in the first 1000 hours of their use. Is there sufficient evidence at the 0.02 level to support the company's claim

The p-value of the test is 0.004 < 0.02, which means that there is sufficient evidence at the 0.02 level to support the company's claim.

Step-by-step explanation:

The company's promotional literature claimed that over 32% do not fail in the first 1000 hours of their use.

At the null hypothesis, we test if the proportion is of at most 32%, that is:

At the alternative hypothesis, we test if the proportion is more than 32%, that is:

The test statistic is:

In which X is the sample mean, is the value tested at the null hypothesis, is the standard deviation and n is the size of the sample.

0.32 is tested at the null hypothesis:

This means that

A sample of 1700 computer chips revealed that 35% of the chips do not fail in the first 1000 hours of their use.

This means that

Value of the test statistic:

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion above 0.35, which is 1 subtracted by the p-value of z = 2.65.

Looking at the z-table, z = 2.65 has a p-value of 0.9960.

1 - 0.9960 = 0.004.

The p-value of the test is 0.004 < 0.02, which means that there is sufficient evidence at the 0.02 level to support the company's claim.

13

step-by-step explanation:

169-1=168

168-3=165

165-5=160

160-7=153

153-9=144

144-11=133

133-13=120

120-15=105

105-17=88

88-19=69

69-21=48

48-23=25

25-25=0

since it took 13times to completely subtract it hence its square root is 13