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test a claim that the mean amount of carbon monoxide in the air in u.s.…

Question

test a claim that the mean amount of carbon monoxide in the air in u.s. cities is less than 2.32 parts per million. it was found that the mean amount of carbon monoxide in the air for the random sample of 65 cities is 2.38 parts per million and the standard deviation is 2.11 parts per million. at \\( \alpha = 0.01 \\), can the claim be supported? complete parts (a) through (e) below. assume the population is normally distributed.
(c) find the standardized test statistic, t.
the standardized test statistic is \\( t = 0.23 \\).
(round to two decimal places as needed.)
(d) decide whether to reject or fail to reject the null hypothesis.
fail to reject \\( h _ { 0 } \\) because the standardized test statistic is not in the rejection region.
(e) interpret the decision in the context of the original claim.
there enough evidence at the % level of significance to the claim that the mean amount of carbon monoxide in the air in u.s. cities is parts per million.
(type integers or decimals. do not round.)

Explanation:

Brief Explanations

Since we failed to reject the null hypothesis \(H_0\), it means that based on the sample data and the significance level \(\alpha = 0.01\), we do not have sufficient evidence to support the alternative claim. The null hypothesis in a one - sample \(t\) - test for a population mean (when the population standard deviation is unknown) is \(H_0:\mu=\mu_0\) (here \(\mu_0 = 2.32\)) and the alternative hypothesis is \(H_1:\mu<2.32\). Failing to reject \(H_0\) implies that the data does not provide strong enough evidence to conclude that the population mean is less than the hypothesized value.

Answer:

There is not enough evidence at the \(1\%\) level of significance to support the claim that the mean amount of carbon monoxide in the air in U.S. cities is less than \(2.32\) parts per million.