Question

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

Answer 1

To determine if the company's claim is supported, a hypothesis test should be performed. A one-sample proportion test can be used to evaluate the evidence. Calculating the test statistic and comparing it to the critical value will determine if there is sufficient evidence to support the company's claim.

Step-by-step explanation:

In order to determine if the company's claim is supported, we need to perform a hypothesis test. Let's assume that the true proportion of computer chips that do not fail in the first 1000 hours is 32%. The null hypothesis (H0) is that the true proportion is equal to 32%, and the alternative hypothesis (H1) is that the true proportion is greater than 32%. We can use a one-sample proportion test to evaluate the evidence.

Using the sample data, we can calculate the test statistic and compare it to the critical value. If the test statistic is greater than the critical value, we can reject the null hypothesis and conclude that there is sufficient evidence to support the company's claim.

To calculate the test statistic, we need to use the formula: z = (p - P0) / sqrt(P0 * (1 - P0) / n), where p is the sample proportion, P0 is the hypothesized proportion, and n is the sample size. If the calculated test statistic is greater than the critical value, we can reject the null hypothesis and conclude that there is sufficient evidence to support the company's claim.

Answer 2

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:

`H_0: p \leq 0.32`

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

`H_1: p > 0.32`

The test statistic is:

`z = (X - \mu)/((\sigma)/(√(n)))`

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

0.32 is tested at the null hypothesis:

This means that
`\mu = 0.32 \sigma = √(0.32*0.68)`

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
`n = 1700, X = 0.35`

Value of the test statistic:

`z = (X - \mu)/((\sigma)/(√(n)))`

`z = (0.35 - 0.32)/((√(0.32*0.68))/(√(1700)))`

`z = 2.65`

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.