Question

A computer manufacturer estimates that its cheapest screens will last less than 2.8 years. A random sample of 61 of these screens has a mean life of 2.7 years. The population is normally distributed with a population standard deviation of 0.88 years. At α=0.02, what type of test is this and can you support the organization’s claim using the test statistic(A) Claim is null, fail to reject the null and cannot support claim as test statistic (-0.89) is not in the rejection region defined by the critical value (-2.33)(B) Claim is alternative, reject the null and support claim as test statistic (-0.89) is not in the rejection region defined by the critical value (-2.33)(C) Claim is null, reject the null and support claim as test statistic (-0.89) is not in the rejection region defined by the critical value (-2.05)(D) Claim is alternative, fail to reject the null and cannot support claim as test statistic (-0.89) is not in the rejection region defined by the critical value (-2.05)

Answer 1

Answer:

This is a left-tailed test about a population's mean claim, with known population standard deviation.

c) Claim is alternative, fail to reject the null and cannot support calim as test statistic (-0.89) is not in the rejection region defined by the critical value (-2.05)

Step-by-step explanation:

The claim is alternative because it does not contain an equality. Since we are testing a claim about a population mean when the population standard deviation is known, the statistic in this case is z.

$z =\frac{x-u}{\frac{s}{\sqrt{n} } } \\=\frac{2.7-2.8}{\frac{0.88}{\sqrt{61} } }$

x is the sample mean, u is the population parameter, s is the population standard deviation and n is the number of items sampled.

z = -0.89

The rejection region is going to be the 0.02 (the alfa level) that is to the left of the probability distribution because this is a left-tailed test since the claim is that the population mean is less than 2.8 years. This region is going to be limited by the critical value -2.05.

In a z-core table we can see that for a z = -2.05 the area to the left of the distribution is 0.0202 ≈ 0.02 which is the significance level.