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claim that the mean amount of lead in the air in u.s. cities is less th…

Question

claim that the mean amount of lead in the air in u.s. cities is less than 0.037 microgram per cubic meter. it was that the mean amount of lead in the air for the random sample of 57 u.s. cities is 0.039 microgram per cubic and the standard deviation is 0.069 microgram per cubic meter. at \\( \alpha = 0.10 \\), can the claim be supported? lete parts (a) through (e) below. assume the population is normally distributed. \\( \mu < 0.037 \\) be integers or decimals. do not round.) e claim is the alternative hypothesis. ) find the critical value(s) and identify the rejection region(s). he critical value(s) is/are \\( t _ { 0 } = - 1.29 \\). use a comma to separate answers as needed. round to two decimal places as needed.) choose the graph which shows the rejection region.

Explanation:

Step1: State the hypotheses

The null hypothesis \(H_0:\mu\geq0.037\) and the alternative hypothesis \(H_1:\mu < 0.037\) (since the claim is that the mean is less than \(0.037\)).

Step2: Calculate the test - statistic

The formula for the \(t\) - test statistic is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\). Here, \(\bar{x} = 0.039\), \(\mu=0.037\), \(s = 0.069\), and \(n = 57\).

$$t=\frac{0.039 - 0.037}{0.069/\sqrt{57}}$$
$$t=\frac{0.002}{0.069/\sqrt{57}}$$
$$t=\frac{0.002}{0.0091}$$
$$t\approx0.22$$

Step3: Make a decision

The critical value \(t_0=- 1.29\). Since the test statistic \(t = 0.22>-1.29\) (the test statistic does not fall in the rejection region \(t < t_0\)), we fail to reject the null hypothesis.

Answer:

We fail to reject the null hypothesis. So, the claim that the mean amount of lead in the air in U.S. cities is less than \(0.037\) microgram per cubic meter is not supported at the \(\alpha = 0.10\) significance level.