Question

a sample of 1600 computer chips revealed that 64% of the chips do not fail in the first 1000 hours of their use. the company's promotional literature claimed that above 61% 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? state the null and alternative hypotheses for the above scenario.

Answer 1

The null hypothesis is that the proportion of computer chips that do not fail in the first 1000 hours is equal to or less than 61%. The alternative hypothesis is that the proportion is greater than 61%. To test these hypotheses, a one-sample proportion test can be performed using the sample proportion and the claimed proportion.

Step-by-step explanation:

The null hypothesis for this scenario is: the proportion of computer chips that do not fail in the first 1000 hours of use is equal to or less than 61%. The alternative hypothesis is: the proportion of computer chips that do not fail in the first 1000 hours of use is greater than 61%.

To test these hypotheses, we can perform a one-sample proportion test. We calculate the test statistic using the sample proportion (0.64) and the claimed proportion (0.61), as well as the sample size (1600) to find the standard error. With this test statistic, we can calculate the p-value.

If the p-value is less than (or equal to) the significance level of 0.02, we reject the null hypothesis and conclude that there is sufficient evidence to support the company's claim. If the p-value is greater than 0.02, we fail to reject the null hypothesis and do not have enough evidence to support the company's claim.

Answer 2

Using a significance level of 0.02, we calculate the test statistic and compare it to the critical value. Since the test statistic is greater than the critical value, we reject the null hypothesis and conclude that there is sufficient evidence to support the company's claim.

Step-by-step explanation:

To determine whether there is sufficient evidence to support the company's claim, we need to evaluate whether the sample proportion of 64% is significantly different from the claimed proportion of above 61%. We can set up the null and alternative hypotheses as follows:

Null hypothesis (H-0): The proportion of chips that do not fail in the first 1000 hours is 61% or less.
Alternative hypothesis (Ha): The proportion of chips that do not fail in the first 1000 hours is greater than 61%.

To test these hypotheses, we can use a one-sample proportion test. We calculate the test statistic using the formula:

Z = (p - p) / sqrt((p * (1 - p)) / n),

where p is the sample proportion, p is the claimed proportion, and n is the sample size. In this case, p = 0.64, p = 0.61, and n = 1600.

Using a significance level of 0.02, we can find the critical value from the standard normal distribution table or using statistical software. For a one-sided test at a 0.02 level of significance, the critical value is approximately 2.055.

Next, we calculate the test statistic:

Z = (0.64 - 0.61) / sqrt((0.61 * (1 - 0.61)) / 1600) ≈ 2.18

Since the test statistic (2.18) is greater than the critical value (2.055), we reject the null hypothesis. Therefore, there is sufficient evidence at the 0.02 level to support the company's claim that above 61% of the chips do not fail in the first 1000 hours of their use.